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# M1-Allens Made Maths Theory + Exercise [II]

## Description: M1-Allens Made Maths Theory + Exercise [II]

JEE-Mathematics EXERCISE - 02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is - (A) x2 + 2y2 – ax = 0 (B) 2x2 + y2 – 2ax = 0 (C) 2x2 + 2y2 – ay = 0 (D) 2x2 + y2 – 2ay = 0 2 . Let A be the vertex and L the length of the latus rectum of parabola, y2 – 2y – 4x – 7 = 0. The equation of the parabola with point A as vertex, 2L as the length of the latus rectum and the axis at right angles to that of the given curve is - (A) x2 + 4x + 8y – 4 = 0 (B) x2 + 4x – 8y + 12 = 0 (C) x2 + 4x + 8y + 12 = 0 (D) x2 + 8x – 4y + 8 = 0 3 . The parametric coordinates of any point on the parabola y2 = 4ax can be - (A) (at2, 2at) (B) (at2, –2at) (C) (asin2t, 2asint) (D) (asint, 2acost) 4 . PQ is a normal chord of the parabola y2 = 4ax at P, A being the vertex of the parabola. Through P a line is drawn parallel to AQ meeting the x-axis in R. Then the length of of AR is - (A) equal to the length of the latus rectum (B) equal to the focal distance of the point P. (C) equal to twice the focal distance of the point P. (D) equal to the distance of the point P from the directrix 5 . The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is- (A) 5 2 (B) 4 5 (C) 4 50 (D) 2 5 6. If the tangents and normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and 7. (x2, y2) respectively, then - 8. (A) x = x (B) x = y (C) y = y (D) x = y 9. 12 12 12 21 10. Locus of the intersection of the tangents at the ends of the normal chords of the parabola y2 = 4ax is - E (A) (2a + x)y2 + 4a3 = 0 (B) (x + 2a)y2 + 4a2 = 0 (C) (y + 2a)x2 + 4a3 = 0 (D) none NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 The locus of the mid point of the focal radii of a variable point moving on the parabola, y2 = 4ax is a parabola whose (A) latus rectum is half the latus rectum of the original parabola (B) vertex is (a/2, 0) (C) directrix is y-axis (D) focus has the co-ordinates (a, 0) The equation of a straight line passing through the point (3, 6) and cutting the curve y = x orthogonally is - (A) 4x + y – 18 = 0 (B) x + y – 9 = 0 (C) 4x – y – 6 = 0 (D) none The tangent and normal at P (t), for all real positive t, to the parabola y2 = 4ax meet the axis of the parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through the points P, T and G is - (A) cot–1t (B) cot–1t2 (C) tan–1t  t (D) sin–1  1  t2  19

JEE-Mathematics 1 1 . A variable circle is described to passes through the point (1, 0) and tangent to the curve y = tan(tan–1x). The locus of the centre of the circle is a parabola whose - (A) length of the latus rectum is 2 2 (B) axis of symmetry has the equation x + y = 1 (C) vertex has the co-ordinates (3/4, 1/4) (D) none of these 1 2 . AB, AC are tangents to a parabola y2 = 4ax. p p and p are the lengths of the perpendiculars from A, 12 3 B and C respectively on any tangent to the curve, then p , p , p are in- 213 (A) A.P. (B) G.P. (C) H.P. (D) none of these 1 3 . Through the vertex O of the parabola, y2 = 4ax two chords OP and OQ are drawn and the circles on OP and OQ as diameter intersect in R. If 1, 2 and  are the angles made with the axis by the tangent at P and Q on the parabola and by OR then the value of cot1 + cot2 = (A) – 2tan (B) – 2tan( – ) (C) 0 (D) 2cot 1 4 . Two parabolas have the same focus. If their directrices are the x-axis & the y-axis respectively, then the slope of their common chord is - (A) 1 (B) –1 (C) 4/3 (D) 3/4 1 5 . Tangent to the parabola y2 = 4ax at point P meets the tangent at vertex A, at point B and the axis of parabola at T. Q is any point on this tangent and N is the foot of perpendicular from Q on SP, where S is focus. M is the foot of perpendicular from Q on the directrix then - (A) B bisects PT (B) B trisects PT (C) QM = SN (D) QM = 2SN 1 6 . If the distance between a tangent to the parabola y2 = 4 x and a parallel normal to the same parabola is 2 2 , then possible values of gradient of either of them are - (A) –1 (B) +1 (C) – 5  2 (D) + 5  2 1 7 . If two distinct chords of a parabola x2 = 4ay passing through (2a, a) are bisected on the line x + y = 1, then length of latus rectum can be - (A) 2 (B) 1 (C) 4 (D) 5 1 8 . If PQ is a chord of parabola x2 = 4y which subtends right angle at vertex. Then locus of centroid of triangle PSQ (S is focus) is a parabola whose - (A) vertex is (0, 3) (B) length of LR is 4/3 (C) axis is x = 0 (D) tangent at the vertex is x = 3 1 9 . Identify the correct statement(s) - (A) In a parabola vertex is the mid point of focus and foot of directrix.    (B) P at12, 2at1 2 are two points on y2 = 4ax such that t1t2 = –1, then normals at P and Q are & Q at 2 , 2at2 perpendicular. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 (C) There doesn't exist any tangent of y2 = 4ax which is parallel to x-axis. (D) At most two normals can be drawn to a parabola from any point on its plane. 2 0 . For parabola y2 = 4ax consider three points A, B, C lying on it. If the centroid of ABC is (h1, k1) & centroid of triangle formed by the point of intersection of tangents at A, B, C has coordinates (h2, k2), then which of the following is always true - 4a  4a 3 3 k 2 h1 2h2   (D)k2 (A) 2k1 = k2 (B) k1 = k2 (C) 1   1 2h1  h2 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B A,B A,B C D C A A,B,C,D A C,D Que. 11 12 13 14 15 16 17 18 19 20 Ans. B,C B A A,B A,C A,B,C,D A,B A,B,C A,B,C B,C 20 E

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Column-I Column-II (p) 4 (A) The normal chord at a point t on the parabola y2 = 4x subtends a right angle at the vertex, then t2 is (q) 2 (B) The area of the triangle inscribed in the curve y2 = 4x. If the parameter of vertices are 1, 2 and 4 is (C) The number of distinct normal possible from  11 , 1  to the (r) 3  4 4  (s) 6 parabola y2 = 4x is (D) The normal at (a, 2a) on y2 = 4ax meets the curve again at (at2, 2at), then the value of |t – 1| is 2 . Column-I Column-II (p) 8 (A) Area of a triangle formed by the tangents drawn from a point (–2, 2) to the parabola y2 = 4(x + y) and their (q) 4 3 corresponding chord of contact is (r) 4 (s) 24/5 (B) Length of the latus rectum of the conic 25{(x – 2)2 + (y – 3)2} = (3x + 4y – 6)2 is (C) If focal distance of a point on the parabola y = x2 – 4 is 25/4 and points are of the form (± a , b) then value of a + b is (D) Length of side of an equilateral triangle inscribed in a parabola y2 – 2x – 2y – 3 = 0 whose one angular point is vertex of the parabola, is ASSERTION & REASON These questions contain, Statement-I (assertion) and Statement-II (reason). 1.NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. 2. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I. (C) Statement-I is true, Statement-II is false. E (D) Statement-I is false, Statement-II is true. Statement-I : If normal at the ends of double ordinate x = 4 of parabola y2 = 4x meet the curve again at P and P' respectively, then PP' = 12 unit. Because Statement-II : If normal at t1 of y2 = 4ax meets the parabola again at t2, then t12 = 2 + t1 t2. (A) A (B) B (C) C (D) D Statement-I : The lines from the vertex to the two extremities of a focal chord of the parabola y2 = 4ax  are at an angle of . 2 Because Statement-II : If extremities of focal chord of parabola are ( at12 , 2at1) and ( at22 , 2at2), then t1t2 = –1 (A) A (B) B (C) C (D) D 21

JEE-Mathematics 3 . Statement-I : If P1Q1 and P2Q2 are two focal chords of the parabola y2 = 4ax, then the locus of point of intersection of chords P1P2 and Q1Q2 is directrix of the parabola. Here P1P2 and Q1Q2 are not parallel. Because Statement-II : The locus of point of intersection of perpendicular tangents of parabola is directrix of parabola. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 Observe the following facts for a parabola : (i) Axis of the parabola is the only line which can be the perpendicular bisector of the two chords of the parabola. (ii) If AB and CD are two parallel chords of the parabola and the normals at A and B intersect at P and the normals at C and D intersect at Q, then PQ is a normal to the parabola. Let a parabola is passing through (0, 1), (–1, 3), (3, 3) & (2, 1) On the basis of above information, answer the following questions : 1 . The vertex of the parabola is - (A) 1, 1 (B)  1 , 1  (C) (1, 3) (D) (3, 1) 3   3  2 . The directrix of the parabola is - 1 1 1 1 (A) y – = 0 (B) y + = 0 (C) y + = 0 (D) y   0 24 2 24 12 3 . For the parabola y2 = 4x, AB and CD are any two parallel chords having slope 1. C1 is a circle passing through O, A and B and C2 is a circle passing through O, C and D, where O is origin. C1 and C2 intersect at - (A) (4, –4) (B) (–4, 4) (C) (4, 4) (D) (–4, –4) Comprehension # 2 : If a source of light is placed at the fixed point of a parabola and if the parabola is a reflecting surface, then the ray will bounce back in a line parallel to the axis of the parabola. On the basis of above information, answer the following questions : 1 . A ray of light is coming along the line y = 2 from the positive direction of x-axis and strikes a concave mirror whose intersection with the xy-plane is a parabola y2 = 8x, then the equation of the reflected ray is - (A) 2x + 5y = 4 (B) 3x + 2y = 6 (C) 4x + 3y = 8 (D) 5x + 4y = 10 2 . A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror whose equation is y2 + 10y – 4x + 17 = 0 After reflection, the ray must pass through the point - (A) (–2, –5) (B) (–1, –5) (C) (–3, –5) (D) (–4, –5) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 3 . Two ray of light coming along the lines y = 1 and y = –2 from the positive direction of x-axis and strikes a concave mirror whose intersection with the xy-plane is a parabola y2 = x at A and B respectively. The reflected rays pass through a fixed point C, then the area of the triangle ABC is - 21 19 17 15 (A) 8 sq. unit (B) sq. unit (C) sq. unit (D) sq. unit 2 2 2 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  Match the Column 1. (A)  (q); (B)  (s); (C)  (q); (D)  (p) 2. (A)  (r); (B)  (s); (C)  (p); (D)  (q)  Assertion & Reason 1. C 2. D 3. B  Comprehension Based Questions Comprehension # 1 : 1. A 2. C 3. A Comprehension # 2 : 1. C 2. B 3. A 22 E

JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the equation of parabola, whose focus is (–3, 0) and directrix is x + 5 = 0. 2 . Find the vertex, axis, focus, directrix, latus rectum of the parabola x2 + 2y – 3x + 5 = 0 3 . Find the equation of the parabola whose focus is (1, –1) and whose vertex is (2, 1). Also find its axis and latus rectum. 4 . If the end points P(t ) and Q( t ) of a chord of a parabola y2 = 4ax satisfy the relation t t = k (constant) then 12 12 prove that the chord always passes through a fixed point. Find that point also ? 5 . Find the locus of the middle points of all chords of the parabola y2 = 4ax which are drawn through the vertex. 6 . O is the vertex of the parabola y² = 4ax & L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H , prove that the length of the double ordinate through H is 4a 5 . 7 . Find the length of the side of an equilateral triangle inscribed in the parabola, y2 = 4x so that one of its angular point is at the vertex. 8 . Two perpendicular chords are drawn from the origin 'O' to the parabola y = x2, which meet the parabola at P and Q. Rectangle POQR is completed. Find the locus of vertex R. 9 . Find the set of values of  in the interval [/2, 3/2], for which the point (sin, cos) does not lie outside the parabola 2y2 + x – 2 = 0. 1 0 . Find the length of the focal chord of the parabola y2 = 4ax whose distance from the vertex is p. 1 1 . If 'm' varies then find the range of c for which the line y = mx + c touches the parabola y2 = 8(x + 2). 1 2 . Find the equations of the tangents to the parabola y² = 16x, which are parallel & perpendicular respectively to the line 2x  y + 5 = 0 . Find also the coordinates of their points of contact . 1 3 . Find the equations of the tangents of the parabola y² = 12x, which passes through the point (2, 5). 1 4 . Prove that the locus of the middle points of all tangents drawn from points on the directrix to the parabola y2 = 4ax is y²(2x + a) = a(3x + a)². NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 5 . Two tangents to the parabola y² = 8x meet the tangent at its vertex in the points P & Q. If PQ = 4 units, prove that the locus of the point of the intersection of the two tangents is y² = 8 (x + 2). 1 6 . Find the equation of the circle which passes through the focus of the parabola x2 = 4y & touches it at the point (6 , 9). 1 7 . In the parabola y² = 4ax, the tangent at the point P, whose abscissa is equal to the latus rectum meets the axis in T & the normal at P cuts the parabola again in Q. Prove that PT : PQ = 4 : 5. 1 8 . Show that the normals at the points (4a , 4a) & at the upper end of the latus rectum of the parabola y² = 4ax intersect on the same parabola. 1 9 . Show that the locus of a point, such that two of the three normals drawn from it to the parabola y² = 4ax are perpendicular is y² = a(x  3a). 2 0 . If the normal at P(18, 12) to the parabola y² = 8x cuts it again at Q, then show that 9PQ = 80 10 2 1 . Prove that the locus of the middle point of portion of a normal to y² = 4ax intercepted between the curve & the axis is another parabola. Find the vertex & the latus rectum of the second parabola. E 23

JEE-Mathematics 2 2 . A variable chord PQ of the parabola y² = 4x is drawn parallel to the line y = x. If the parameters of the points P & Q on the parabola are p & q respectively, show that p + q = 2. Also show that the locus of the point of intersection of the normals at P & Q is 2x  y = 12. 2 3 . P & Q are the points of contact of the tangents drawn from the point T to the parabola y² = 4ax. If PQ be the normal to the parabola at P, prove that TP is bisected by the directrix. 2 4 . The normal at a point P to the parabola y² = 4ax meets its axis at G. Q is another point on the parabola such that QG is perpendicular to the axis of the parabola. Prove that QG²  PG² = constant. 2 5 . Three normals to y² = 4x pass through the point (15, 12). Show that one of the normals is given by y = x  3 & find the equations of the others. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 . y2 = 4(x + 4) 2. Vertex   3 , 1 1  , focus   3 , 15  , 37  2 8   2 8  axis : x = , directrix : y = – , latus rectum = 2 28 3 . (2x – y – 3)2 = –20(x + 2y – 4), axis : 2x – y – 3 = 0. latus rectum = 4 5 . 4 . (–ak, 0) 5 . y2 = 2ax 7 . 8 3 8 . x2 = y – 2 9 .  [ / 2, 5 / 6]  [, 3 / 2] 4a3 1 1 . (–, –4]  [4, ) 1 0 . p2 1 2 . 2x  y + 2 = 0, (1, 4) ; x + 2y + 16 = 0, (16, 16) 1 3 . 3x  2y + 4 = 0 ; x  y + 3 = 0 1 6 . x2 + y2 + 18 x  28 y + 27 = 0 2 1 . (a , 0) ; a 2 5 . y = 4x + 72 , y = 3x  33 24 E

EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 . If from the vertex of a parabola a pair of chords be drawn at right angles to one another, & with these chords as adjacent sides a rectangle be constructed , then find the locus of the outer corner of the rectangle. 2 . Two perpendicular straight lines through the focus of the parabola y² = 4ax meet its directrix in T & T respectively. Show that the tangents to the parabola parallel to the perpendicular lines intersect in the mid point of T T '. 3 . Find the condition on ‘a’ & ‘b’ so that the two tangents drawn to the parabola y² = 4ax from a point are normals to the parabola x² = 4by. 4 . TP & TQ are tangents to the parabola and the normals at P & Q meet at a point R on the curve. Prove that the centre of the circle circumscribing the triangle TPQ lies on the parabola 2 y² = a(x  a). 5 . Let S is the focus of the parabola y2 = 4ax and X the foot of the directrix, PP' is a double ordinate of the curve and PX meets the curve again in Q. Prove that P'Q passes through focus. 6 . Prove that on the axis of any parabola y² = 4ax there is a certain point K which has the property that , if a chord 11 PQ of the parabola be drawn through it , then  is same for all positions of the chord. Find P K 2 Q K 2 also the coordinates of the point K. 7 . If (x1, y1), (x2, y2) and (x3, y3) be three points on the parabola y2 = 4ax and the normals at these points meet in a point, then prove that x1  x2 + x2  x3 + x3  x1 = 0 y3 y1 y2 8 . A variable chord joining points P(t ) and Q(t ) of the parabola y² = 4ax subtends a right angle at a fixed point 12 t of the curve. Show that it passes through a fixed point. Also find the co-ordinates of the fixed point. 0 9 . Show that a circle circumscribing the triangle formed by three co-normal points passes through the vertex of the parabola and its equation is, 2(x2+y2) – 2(h+2a) x – ky = 0, where (h, k) is the point from where three concurrent normals are drawn. 1 0 . A ray of light is coming along the line y = b from the positive direction of x-axis & strikes a concave mirror whose intersection with the xy-plane is a parabola y2 = 4 ax. Find the equation of the reflected ray & show that it passes through the focus of the parabola. Both a & b are positive. [REE 95] BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . y² = 4a(x  8a) 3 . a² > 8b² 6. (2a , 0) 8 . [a(t² + 4),  2at ] 10. 4abx + (4a² – b²)y – 4a²b = 0 00 E 25

JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . The length of the latus rectum of the parabola x2 – 4x – 8y + 12 = 0 is- [AIEEE-2002] (1) 4 (2) 6 (3) 8 (4) 10 2 . The equation of tangents to the parabola y2 = 4ax at the ends of its latus rectum is- [AIEEE-2002] (1) x – y + a = 0 (2) x + y + a = 0 (3) x + y – a = 0 (4) both (1) and (2) 3 . The normal at the point (bt12, 2bt1) on a parabola meets the parabola again in the point (bt22, 2bt2), then- [AIEEE-2003] 2 2 2 2 (1) t = t + (2) t = –t – (3) t = –t + (4) t = t – 2 1 t1 2 1 t1 2 1 t1 2 1 t1 4 . If a  0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2 = 4ax and x2 = 4ay, then- [AIEEE-2004] (1) d2 + (2b + 3c)2 = 0 (2) d2 + (3b + 2c)2 = 0 (3) d2 + (2b – 3c)2 = 0 (4) d2 + (3b – 2c)2 = 0 a3 x2 a2x 5 . The locus of the vertices of the family of parabolas y = + – 2a is- [AIEEE-2006] 32 3 35 64 105 (1) xy = (2) xy = 16 (3) xy = 105 (4) xy = 64 4 6 . The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangents is- [AIEEE-2007] (1) (–1, 1) (2) (0, 2) (3) (2, 4) (4) (–2, 0) 7 . A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at - [AIEEE-2008] (1) (0, 2) (2) (1, 0) (3) (0, 1) (4) (2, 0) 8 . If two tangents drawn from a point P to the parabola y2 = 4x are at right angles then the locus of P is :- [AIEEE-2010] (1) x = 1 (2) 2x + 1 = 0 (3) x = –1 (4) 2x – 1 = 0 9 . Given : A circle, 2x2 + 2y2 = 5 and a parabola, y2 = 4 5 x. [JEE (Main)-2013] Statement–I : An equation of a common tangent to these curves is y = x + 5 . 5 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 Statement–II : If the line, y = mx + (m  0) is their common tangent, then m satisfies m4 – 3m2 + 2 = 0. m (1) Statement-I is true, Statement-II is true; statement-II is a correct explanation for Statement-I. (2) Statement-I is true, Statement-II is true; statement-II is not a correct explanation for Statement-I. (3) Statement-I is true, Statement-II is false. (4) Statement-I is false, Statement-II is true. PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Qu e. 1 2 3 4 5 6 7 8 9 Ans 3 4 2 1 4 4 2 3 2 E 26

JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . ( a ) If the line x – 1 = 0 is the directrix of the parabola y2 – kx + 8 = 0, then one of the values of ‘k’ is: (A) 1/8 (B) 8 (C) 4 (D) 1/4 ( b ) If x + y = k is normal to y2 = 12x , then ‘k’ is - [JEE 2000 ( Screening) 1+1M] (A) 3 (B) 9 (C) – 9 (D) – 3 2 . ( a ) The equation of the common tangent touching the circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above the x- axis is - (A) 3y  3x  1 (B) 3y  (x  3) (C) 3y  x  3 (D) 3y  (3 x  1) [JEE 2001 ( Screening) 1+1M] ( b ) The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is - (A) x = – 1 (B) x = 1 (C) x = –3/2 (D) x = 3/2 3 . The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with directrix [JEE 2002 ( Screening), 3M] (A) x = – a a (C) x = 0 a (B) x = – (D) x = 2 2 [JEE 2002 (Scr), 3M] 4 . The equation of the common tangent to the curves y2 = 8x and xy = – 1 is - (A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2 5 . If a focal chord of the parabola y 2 = 16x is a tangent to the circle (x – 6)2 + y2 = 2. then the set of possible values of the slope of this chord, are - [JEE 2003 (Scr), 3M] 2 , 1 2 ,  1  (A) {–1, 1} (B) {–2, 2} (C)   (D)  2  2   6 . Normals with slopes m , m , m are drawn from the point P to the parabola y2 = 4x. If locus of P with 123 m m =  is a part of the parabola itself, find . [JEE 2004 (Mains), 4M out of 60] 1 2 7 . Two tangents are drawn from point (1, 4) to the parabola y2 = 4x. Angles between tangents is - (A) /6 (B) /4 (C) /3 (D) /2 [JEE 2004 (Screening), 3M] 8 . At any point P on the parabola y2 – 2y – 4x + 5 = 0, a tangent is drawn which meets the directrix at Q. Find the 1 [JEE 2004 (Mains), 4M out of 60] locus of point R which divides QP externally in the ratio : 1 . 2 9 . Tangent to the curve y = x2 + 6 at point P (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then coordinate of Q is - [JEE 2005 (Screening) 3M] (A) (–6, 11) (B) (6, –11) (C) (–6, –7) (D) (–6, –11) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 1 0 . The axis of a parabola is along the line y = x and the distance of its vertex from origin is 2 and that of origin from its focus is 2 2 . If vertex and focus both lie in the first quadrant, then the equation of the parabola is - [JEE 2006 (3M, –1M) out of 184] (A) (x + y)2 = (x – y – 2) (B) (x – y)2 = (x + y – 2) (C) (x – y)2 = 4(x + y – 2) (D) (x – y)2 = 8(x + y – 2) 1 1 . The equations of the common tangents to the parabola y = x2 and y = – x2 + 4x – 4 is/are- (A) y = 4(x – 1) (B) y = 0 (C) y = –4(x – 1) (D) y = –30x – 50 [JEE 2006 , (5M, –1M) out of 184] 12. Match the following [JEE 2006, (6M, 0M) out of 184] Normals are drawn at points P, Q and R lying on the parabola y2 = 4x which intersect at (3, 0). Then (i) Area of PQR (A) 2 (ii) Radius of circumcircle of PQR (B) 5/2 (iii) Centroid of PQR (C) (5/2, 0) (iv) Circumcentre of PQR (D) (2/3, 0) E 27

JEE-Mathematics 13 to 15 are based on this paragraph [JEE 2006 (5M, –2M) each, out of 184] Let ABCD be a square of side length 2 units. C2 is the circle through vertices A, B, C, D and C1 is the circle touching all the sides of the square ABCD. L is a line through A. PA 2  PB2  PC2  PD2 1 3 . If P is a point on C1 and Q in another point on C2, then QA 2  QB 2  QC 2  QD2 is equal to - (A) 0.75 (B) 1.25 (C) 1 (D) 0.5 1 4 . A circle touches the line L and circle C1 externally such that both the circles are on the same side of the line, then the locus of centre of the circle is - (A) ellipse (B) hyperbola (C) parabola (D) pair of straight line 1 5 . A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts M at T2 and T3 and AC at T1 then area of T1T2T3 is (A) 1/2 sq. units (B) 2/3 sq. units (C) 1 sq. units (D) 2 sq. units 16 to 18 are based on this paragraph Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. 1 6 . The ratio of the areas of the triangle PQS and PQR is :- [JEE 2007, 4M] (A) 1 : 2 (B) 1 : 2 (C) 1 : 4 (D) 1 : 8 [JEE 2007, 4M] 1 7 . The radius of the circumcircle of the triangle PRS is :- (D) 2 3 (A) 2 (B) 3 3 (C) 3 2 [JEE 2007, 4M] 1 8 . The radius of the incircle of the triangle PQR is :- (A) 4 (B) 3 8 (D) 2 Assertion and Reason : (C) 3 1 9 . Statement-1 : The curve y = x2 + x + 1 is symmetric with respect to the line x = 1 because 2 Statement-2 : A parabola is symmetric about its axis. [JEE 2007, 3M] (A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 2 0 . Consider the two curves C1 : y2 = 4x ; C2 : x2 + y2 – 6x + 1 = 0. Then [JEE 2008, 3M, –1M] (A) C1 and C2 touch each other only at one point (B) C1 and C2 touch each other exactly at two points (C) C1 and C2 intersect (but do not touch) at exactly two points (D) C1 and C2 neither intersect nor touch each other 2 1 . The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose [JEE 2009, 4M, –1M] (A) vertex is  2a , 0 (B) directrix is x =0 2a (D) focus is (a, 0)  3 (C) latus rectum is 3 2 2 . Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be - [JEE 2010, 3M] (A) –1/r (B) 1/r (C) 2/r (D) –2/r 2 3 . Consider the parabola y2 = 8x. Let 1 be the area of the triangle formed by the end points of its latus rectum and the point P  1 , 2  on the parabola, and 2 be the area of the triangle formed by drawing tangents  2  at P and at the end points of the latus rectum. Then 1 is [JEE 2011,4M] 2 E 28

JEE-Mathematics 2 4 . Let (x,y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1 : 3. Then the locus of P is - [JEE 2011,3M] (A) x2 = y (B) y2 = 2x (C) y2 = x (D) x2 = 2y 2 5 . Let L be a normal to the parabola y2 = 4x. If L passes through the point (9,6), then L is given by - [JEE 2011,4M] (A) y – x + 3 =0 (B) y + 3x – 33 = 0 (C) y + x – 15 = 0 (D) y – 2x + 12 = 0 2 6 . Let S be the focus of the parabola y2 = 8x & let PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of the triangle PQS is [JEE 2012, 4M] Paragraph for Question 27 and 28 Let PQ be a focal chord of the parabolas y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0. 2 7 . If chord PQ subtends an angle  at the vertex of y2 = 4ax, then tan = [JEE(Advanced) 2013, 3, (–1)M] 2 7 2 7 2 5 2 5 (A) 3 (B) 3 (C) 3 (D) 3 2 8 . Length of chord PQ is [JEE(Advanced) 2013, 3, (–1)M] (A) 7a (B) 5a (C) 2a (D) 3a 2 9 . A line L : y = mx + 3 meets y-axis at E(0,3) and the arc of the parabola y2 = 16x, 0 < y < 6 at the point F(x ,y ). The tangent to the parabola at F(x ,y ) intersects the y-axis at G(0,y ). The slope m of the line L 00 00 1 is chosen such that the area of the triangle EFG has a local maximum. Match List-I with List-II and select the correct answer using the code given below the lists. List-I List-II NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#09\\Eng\\01 PARABOLA.p65 P. m = 1 Q. Maximum area of EFG is 1. R. y = 2 0 2. 4 S. y = 3. 2 1 4. 1 Codes : [JEE(Advanced) 2013, 3, (–1)M] P QR S (A) 4 1 2 3 (B) 3 4 1 2 (C) 1 3 2 4 (D) 1 3 4 2 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1. (a) C (b) B 2. (a) C (b) D 3. C 4. D 5. A 6. 2 7 . C 8 . (x + 1)(y – 1)2 + 4 = 0 9. C 16. C 10. D 1 1 . A, B 12. (i) A, (ii) B, (iii) D, (iv) C 13. A 14. C 15. C 24. C 17. B 18. D 19. A 20. B 2 1 . A,D 2 2 . C, D 23. 2 25. A,B,D 26. 4 27. D 28. B 29. A E 29

JEE-Mathematics PERMUTATION & COMBINATION 1 . FUNDAMENTAL PRINCIPLE OF COUNTING (counting without actual counting): If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total number of different ways of- (a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number of events (known as multiplication principle). (b) happening exactly one of the events is m + n (known as addition principle). Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT and branch in 15 × 10 = 150 number of ways. Example : There are 15 IITs & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can select an institute in (15 + 20) = 35 number of ways. Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is- (A) 24 (B) 2 (C) 12 (D) 10 Solution : The student has 6 choices from the morning courses out of which he can select one course in 6 ways. For the evening course, he has 4 choices out of which he can select one in 4 ways. Hence the total number of ways 6 × 4 = 24. Ans.(A) Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening- (A) 6 (B) 4 (C) 10 (D) 24 Solution : The student has 6 choices from the morning courses out of which he can select one course in 6 ways. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 For the evening course, he has 4 choices out of which he can select one in 4 ways. Hence the total number of ways 6 + 4 = 10. Ans. (C) Do yourself - 1 : ( i ) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many ways can a person travel from A to C ? ( i i ) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a person select two balls ? 2. PERMUTATION & COMBINATION : ( a ) Factorial : A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1; E n! = n. (n – 1)! where n  N Note : (i ) 0! = 1! = 1 (ii) Factorials of negative integers are not defined. (i i i ) n! is also denoted by n 1

JEE-Mathematics ( i v ) (2n)! = 2n.n! [1. 3. 5. 7........(2n – 1)] ( v ) Prime factorisation of n! : Let p be a prime number and n be a positive integer, then exponent of p in n! is denoted by Ep (n!) and is given by n  n   n  n  Ep(n!) =  p  +  p 2  +  p3  + ..... +  pk          where pk < n < pk+1 and [x] denotes the integral part of x. If we isolate the power of each prime contained in any number n, then n can be written as n = 21 · 32 · 53 · 74 .... where i are whole numbers. ( b ) Permutation : Each of the arrangements in a definite order which can be made by taking some or all of the things at a time is called a PERMUTATION. In permutation, order of appearance of things is taken into account; when the order is changed, a different permutation is obtained. Generally, it involves the problems of arrangements (standing in a line, seated in a row), problems on digit, problems on letters from a word etc. nPr denotes the number of permutations of n different things, taken r at a time (n  N, r  W, r  n) n! nPr = n (n – 1) (n – 2) ............. (n – r + 1) = (n  r)! Note : ( i ) nPn = n!, nP0= 1, nP1= n (ii) Number of arrangements of n distinct things taken all at a time = n! (i i i ) nPr is also denoted by A n or P(n,r). r (c) Combination : Each of the groups or selections which can be made by taking some or all of the things without considering the order of the things in each group is called a COMBINATION. Generally, involves the problem of selections, choosing, distributed groups formation, committee formation, geometrical problems etc. nCr denotes the number of combinations of n different things taken r at a time (n  N, r  W, r < n) n Cr  n! NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 r!(n  r)! Note : (i) nCr is also denoted by n or C (n, r).  r  ( i i ) nPr = nCr. r! Illustration 3: Find the exponent of 6 in 50! Solution: E2 (50!)  50   50   50   50   50   50  (where [ ] denotes integral part)  2   4   8  16   3 2   6 4  E2(50!) = 25 + 12 + 6 + 3 + 1 + 0 = 47 E3(50!) = 50    50    50    50   3   9   27   81  E3(50!) = 16 + 5 + 1 + 0 = 22  50! can be written as 50! = 247. 322......... Therefore exponent of 6 in 50! = 22 Ans. 2 E

JEE-Mathematics Illustration 4 : If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of (x – 11) things taken all at a time such that a = 182 bc, then the value of x is (A) 15 (B) 12 (C) 10 (D) 18 Solution :  Px2 x 2  a a  x2 ! x P11  b  b  x x!  11!  and Px 11  c  c  x 11 ! x 11  a  182bc  x  2 !  182  x x!  x  11!  x  2x  1  182  14 13  11!  x 1  13  x  12 Ans. (B) Illustration 5 : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn Solution : so that there are atleast two balls of each colour ? The selections of 6 balls, consisting of atleast two balls of each colour from 5 red and 6 white balls, can be made in the following ways Red balls (5) White balls (6) Number of ways 2 4 3 3 5 C2  6C4  150 4 2 5 C3  6C3  200 5 C4  6C2  75 Therefore total number of ways = 425 Ans. Illustration 6 : How many 4 letter words can be formed from the letters of the word 'ANSWER' ? How many of these words start with a vowel ? Solution : Number of ways of arranging 4 different letters from 6 different letters are 6 C4 4!  6!  360 . 2! There are two vowels (A & E) in the word 'ANSWER'. Total number of 4 letter words starting with A :A ___= 5 C3 3!  5!  60 2! NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Total number of 4 letter words starting with E : E _ _ _ = 5 C3 3!  5!  60 2!  Total number of 4 letter words starting with a vowel = 60 + 60 = 120. Ans. Illustration 7 : If all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary, then find the rank of the word 'RAPID'. Solution : First of all, arrange all letters of given word alphabetically : 'ADIPR' Total number of words starting with A _ _ _ _ = 4! = 24 Total number of words starting with D _ _ _ _ = 4! = 24 Total number of words starting with I _ _ _ _ = 4! = 24 Total number of words starting with P _ _ _ _ = 4! = 24 Total number of words starting with RAD _ _ = 2! = 2 Total number of words starting with RAI _ _ = 2! = 2 Total number of words starting with RAPD _ =1 Total number of words starting with RAPI _ =1  Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102 Ans. E3

JEE-Mathematics Do yourself -2 : ( i ) Find the exponent of 10 in 75C25. ( i i ) If 10Pr = 5040, then find the value of r. (i i i ) Find the number of ways of selecting 4 even numbers from the set of first 100 natural numbers. ( iv) If all letters of the word 'RANK' are arranged in all possible manner as they are in a dictionary, then find the rank of the word 'RANK'. ( v ) How many words can be formed using all letters of the word 'LEARN' ? In how many of these words vowels are together ? 3 . PROPERTIES OF nPr and nCr : ( a ) The number of permutation of n different objects taken r at a time, when p particular objects are always to be included is r!.n–pCr–p (p  r  n) ( b ) The number of permutations of n different objects taken r at a time, when repetition is allowed any number of times is nr. ( c ) Following properties of nCr should be remembered : (i) nCr = nCn–r ; nC0 = nCn = 1 (ii) nCx = nCy  x = y or x + y = n (iii) nCr + nCr–1 = n+1Cr (iv) nC0 + nC1 + nC2 + ............ + nCn = 2n (v) nCr = n n–1Cr–1 r (vi) nCr is maximum when n if n is even & r  n  1 or r  n  1 if n is odd. r 22 2 ( d ) The number of combinations of n different things taking r at a time, (i) when p particular things are always to be included = n – pCr–p (ii) when p particular things are always to be excluded = n – pCr (iii) when p particular things are always to be included and q particular things are to be excluded  n – p – qCr–p Illustration 8 : There are 6 pockets in the coat of a person. In how many ways can he put 4 pens in these Solution : Illustration 9 : pockets? Solution : (A) 360 (B) 1296 (C) 4096 (D) none of these First pen can be put in 6 ways. Similarly each of second, third and fourth pen can be put in 6 ways. Hence total number of ways = 6 × 6 × 6 × 6 = 1296 Ans.(B) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 A delegation of four students is to be selected from a total of 12 students. In how many ways can the delegation be selected, if- (a) all the students are equally willing ? (b) two particular students have to be included in the delegation ? (c) two particular students do not wish to be together in the delegation ? (d) two particular students wish to be included together only ? (e) two particular students refuse to be together and two other particular students wish to be together only in the delegation ? (a) Formation of delegation means selection of 4 out of 12. Hence the number of ways = 12C4 = 495. (b) If two particular students are already selected. Here we need to select only 2 out of the remaining 10. Hence the number of ways = 10C2 = 45. (c) The number of ways in which both are selected = 45. Hence the number of ways in which the two are not included together = 495 – 45 = 450 (d) There are two possible cases (i) Either both are selected. In this case, the number of ways in which the selection can be made = 45. (ii) Or both are not selected. In this case all the four students are selected from the remaining ten students. This can be done in 10C4 = 210 ways. Hence the total number of ways of selection = 45 + 210 = 255 4E

JEE-Mathematics (e) We assume that students A and B wish to be selected together and students C and D do not wish to be together. Now there are following 6 cases. (i) (A, B, C) selected, (D) not selected (ii) (A, B, D) selected, (C) not selected (iii) (A, B) selected, (C, D) not selected (iv) (C) selected, (A, B, D) not selected (v) (D) selected, (A, B, C) not selected (vi) A, B, C, D not selected Ans. For (i) the number of ways of selection = 8C1 = 8 For (ii) the number of ways of selection = 8C1 = 8 For (iii) the number of ways of selection = 8C2 = 28 For (iv) the number of ways of selection = 8C3 = 56 For (v) the number of ways of selection = 8C3 = 56 For (vi) the number of ways of selection = 8C4 = 70 Hence total number of ways = 8 + 8 + 28 + 56 + 56 + 70 = 226. Illustration 10: In the given figure of squares, 6 A's should be written in such a Solution : manner that every row contains at least one 'A'. In how many number of ways is it possible ? (A) 24 (B) 25 (C) 26 (D) 27 There are 8 squares and 6 'A' in given figure. First we can put 6 'A' in these 8 squares by 8C6 number of ways. (I) A A A A AA AA (II) A A A A According to question, atleast one 'A' should be included in each row. So after subtracting these two cases, number of ways are = (8C6 – 2) = 28 – 2 = 26. Ans. (C) Illustration 11: There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum Solution : number of triangles with vertices at these points is : Illustration NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Solution : (A) 3p2 (p – 1) + 1 (B) 3p2 (p – 1) (C) p2 (4p – 3) (D) none of these The number of triangles with vertices on different lines = pC1 × pC1 × pC1 = p3 The number of triangles with two vertices on one line and the third vertex on any one of the other two lines = 3C1 {pC2 × 2pC1} = p(p 1) 6p. 2 So, the required number of triangles = p3 + 3p2 (p – 1) = p2 (4p – 3) Ans. (C) 12: There are 10 points in a row. In how many ways can 4 points be selected such that no two of them are consecutive ? Total number of remaining non-selected points = 6 ...... Total number of gaps made by these 6 points = 6 + 1 = 7 If we select 4 gaps out of these 7 gaps and put 4 points in selected gaps then the new points will represent 4 points such that no two of them are consecutive. x. . x. x. . x. Total number of ways of selecting 4 gaps out of 7 gaps = 7C4 Ans. In general, total number of ways of selection of r points out of n points in a row such that no two of them are consecutive : n–r+1Cr E5

JEE-Mathematics Do yourself-3 : ( i ) Find the number of ways of selecting 5 members from a committee of 5 men & 2 women such that all women are always included. ( i i ) Out of first 20 natural numbers, 3 numbers are selected such that there is exactly one even number. How many different selections can be made ? (i i i ) How many four letter words can be made from the letters of the word 'PROBLEM'. How many of these start as well as end with a vowel ? 4 . FORMATION OF GROUPS : ( a ) (i) The number of ways in which (m + n) different things can be divided into two groups such that one of (m  n)! them contains m things and other has n things, is m ! n ! (m  n). (ii) If m = n, it means the groups are equal & in this case the number of divisions is (2n)! . As in any n! n! 2! one way it is possible to interchange the two groups without obtaining a new distribution. (iii) If 2n things are to be divided equally between two persons then the number of ways : (2n)! 2! . n! n! (2!) ( b ) (i) Number of ways in which (m + n + p) different things can be divided into three groups containing m, n & p things respectively is : (m  n  p)! , m  n  p. m! n! p! (ii) If m = n = p then the number of groups = (3n)! . n! n! n! 3! (iii) If 3n things are to be divided equally among three people then the number of ways in which it can be (3n)! done is (n !)3 . ( c ) In general, the number of ways of dividing n distinct objects into  groups containing p objects each and m n!(  m)! groups containing q objects each is equal to p ! q !m !m ! Here p + mq = n Illustration 13 : In how many ways can 15 students be divided into 3 groups of 5 students each such that 2 particular students are always together ? Also find the number of ways if these groups are to be sent to three different colleges. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Solution : Assuming two particular students as one student (as they are always together), we have to make groups of 5 + 5 + 4 students out of 14 students. 14! Therefore total number of ways = 5 !5 !4 !2 ! Now if these groups are to be sent to three different colleges, the total number of ways = 14! 3! Ans. 5 !5 !4 !2 ! Illustration 14 : Find the number of ways of dividing 52 cards among 4 players equally such that each gets exactly one Ace. 48 ! Solution : Total number of ways of dividing 48 cards (Excluding 4Aces) in 4 groups  (12 !)4 4 ! 48 ! Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = 4! (12 !)4 4 ! Now, distribute these groups of cards among four players 48! 48!  4!4!  4! Ans. (12 !)4 4 ! (12 !)4 6E

JEE-Mathematics Illustration 15 : In how many ways can 8 different books be distributed among 3 students if each receives at least 2 books ? Solution : If each receives at least two books, then the division trees would be as shown below : 88 224 332 (i) (ii)  8!  The number of ways of division for tree in figure (i) is (2 !)2 4 !2 !  .  8!  The number of ways of division for tree in figure (ii) is  (3 !)2 2 !2 ! . The total number of ways of distribution of these groups among 3 students is  (2 8! !2!  (3 8! !2   3! . Ans.  !)2 4 !)2 2 !  Do yourself-4 : ( i ) Find the number of ways in which 16 constables can be assigned to patrol 8 villages, 2 for each. (ii) In how many ways can 6 different books be distributed among 3 students such that none gets equal number of books ? (i i i ) n different toys are to be distributed among n children. Find the number of ways in which these toys can be distributed so that exactly one child gets no toy. 5 . PRINCIPLE OF INCLUSION AND EXCLUSION : U In the Venn's diagram (i), we get AB n(A  B) = n(A) + n(B) – n(A B) n(A'  B') = n(U) – n(A  B) (i) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 In the Venn's diagram (ii), we get A U n(A  B C) = n(A) + n(B) + n(C) – n(A B) – n(B C) – n(A  C) + n(A  B  C) B n(A'  B' C') = n(U) – n(A B C) In general, we have n(A1 A2 ........ An)     n(A i )  n(A i  A j )  n(A i  A j  A k )  .....  (1)n n(A1  A 2  ...  A n ) C ij ijk (ii) Illustration 16 : Find the number of permutations of letters a,b,c,d,e,f,g taken all at a time if neither 'beg' nor 'cad' pattern appear. A BU Solution : The total number of permutations without any restrictions; n(U) = 7! beg acdf Let A be the set of all possible permutations in which 'beg' pattern always appears : n(A) = 5! cad befg Let B be the set of all possible permutations in which 'cad' pattern always appears : n(B) = 5! cad beg f E7

JEE-Mathematics n(A  B) : Number of all possible permutations when both 'beg' and 'cad' patterns appear. n(A  B) = 3!. Therefore, the total number of permutations in which 'beg' and 'cad' patterns do not appear n(A'  B') = n(U) – n(A B) = n(U) – n(A) – n(B) + n(AB) = 7! – 5! – 5! + 3!. Ans. Do yourself-5 : ( i ) Find the number of n digit numbers formed using first 5 natural numbers, which contain the digits 2 & 4 essentially. 6 . PERMUTATIONS OF ALIKE OBJECTS : Case-I : Taken all at a time - The number of permutations of n things taken all at a time : when p of them are similar of one type, q of them are similar of second type, r of them are similar of third type and the remaining n – (p + q+ r) are all different n! is : . p! q! r! Illustration 17 : In how many ways the letters of the word \"ARRANGE\" can be arranged without altering the relative position of vowels & consonants. Solution : 4! The consonants in their positions can be arranged in = 12 ways. 2! 3! The vowels in their positions can be arranged in = 3 ways 2!  Total number of arrangements = 12 × 3 = 36 Ans. Illustration 18 : How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places? (A) 17 (B) 18 (C) 19 (D) 20 Solution : There are 4 odd digits (1, 1, 3, 3) and 4 odd places (first, third, fifth and seventh). At these places 4! the odd digits can be arranged in 6 ways 2!2! 3! NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Then at the remaining 3 places, the remaining three digits (2, 2, 4) can be arranged in  3 2! ways  The required number of numbers = 6 × 3 = 18. Ans. (B) Illustration 19 : (a) How many permutations can be made by using all the letters of the word HINDUSTAN ? (b) How many of these permutations begin and end with a vowel ? (c) In how many of these permutations, all the vowels come together ? (d) In how many of these permutations, none of the vowels come together ? (e) In how many of these permutations, do the vowels and the consonants occupy the same Solution : relative positions as in HINDUSTAN ? (a) The total number of permutations = Arrangements of nine letters taken all at a time 9! = = 181440. 2! (b) We have 3 vowels and 6 consonants, in which 2 consonants are alike. The first place can be 7! filled in 3 ways and the last in 2 ways. The rest of the places can be filled in ways. 2! 7! E Hence the total number of permutations = 3 × 2 × = 15120. 2! 8

JEE-Mathematics (c) Assume the vowels (I, U, A) as a single letter. The letters (IUA), H, D, S, T, N, N can be 7! arranged in ways. Also IUA can be arranged among themselves in 3! = 6 ways. 2! 7! Hence the total number of permutations = 2 ! × 6 = 15120. (d) Let us divide the task into two parts. In the first, we arrange the 6 consonants as shown 6! below in ways. 2! × C × C × C × C× C × C × (Here C stands for a consonant and × stands for a gap between two consonants) Now 3 vowels can be placed in 7 places (gaps between the consonants) in 7C3.3! = 210 ways. 6! Hence the total number of permutations = × 210 = 75600. 2! (e) In this case, the vowels can be arranged among themselves in 3! = 6 ways. 6! Also, the consonants can be arranged among themselves in ways. 2! 6! Ans. Hence the total number of permutations = × 6 = 2160. 2! Illustration 20 : If all the letters of the word 'PROPER' are arranged in all possible manner as they are in a dictionary, then find the rank of the word 'PROPER' . Solution : First of all, arrange all letters of given word alphabetically : EOPPRR Total number of words starting with- E_____ = 5!  30 2!2! O_____ = 5!  30 2!2! PE _ _ _ _ = 4!  12 2! PO _ _ _ _ = 4!  12 2! PP _ _ _ _ = 4!  12 2! NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 PRE _ _ _ = 3! = 6 PROE _ _ = 2! = 2 PROPER = 1= 1 Rank of the word PROPER = 105 Ans. Case-II : Taken some at a t ime Illustration 21 : Find the total number of 4 letter words formed using four letters from the word ''PARALLELOPIPED'. Solution : Given letters are PPP, LLL, AA, EE, R, O, I, D. Cases No.of ways No.of ways Total All distinct of selection 1680 2 alike, 2 distinct of arrangements 1008 2 alike, 2 other alike 8 C4 3 alike, 1 distinct 4 C1  7C2 8C4 4! 36 4 C2 4 C1  7C2  4! 56 2! 2780 2 C1  7C1 4 C2  4! 2!2! 2 C1  7C1  4! 3! Total Ans. E9

JEE-Mathematics Illustration 22 : Find the number of all 6 digit numbers such that all the digits of each number are selected from the set {1,2,3,4,5} and any digit that appears in the number appears at least twice. Solution : Cases No.of ways No.of ways Total All alike of selection 5 4 alike 2 other alike of arrangements 5 C1 300 3 alike  3 other alike 5 C2 2! 5 C1 1 200 2 alike  2 other alike 2 other alike 5 C2 5 C2  2  6! 900 2!4 1405 5 C3 ! 5 C2  6! 3!3! 5 C3  6! 2!2!2! Total Ans. Do yourself-6 : ( i ) In how many ways can the letters of the word 'ALLEN' be arranged ? Also find its rank if all these words are arranged as they are in dictionary. (ii) How many numbers greater than 1000 can be formed from the digits 1, 1, 2, 2, 3 ? 7 . CIRCULAR PERMUTATION : CBAD D BC AB DA C ADCB ABCD DABC CDAB BCDA (a) (b) (c) (d) Let us consider that persons A,B,C,D are sitting around a round table. If all of them (A,B,C,D) are shifted by one place in anticlockwise order, then we will get Fig.(b) from Fig.(a). Now, if we shift A,B,C,D in anticlockwise order, we will get Fig.(c). Again, if we shift them, we will get Fig.(d) and in the next time, Fig.(a). Thus, we see that if 4 persons are sitting at a round table, they can be shifted four times and the four different arrangements, thus obtained will be the same, because anticlockwise order of A,B,C,D does not change. But if A,B,C,D are sitting in a row and they are shifted in such an order that the last occupies the place of first, then the four arrangements will be different. Thus, if there are 4 things, then for each circular arrangement number of linear arrangements is 4. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Similarly, if n different things are arranged along a circle, for each circular arrangement number of linear arrangements is n. Therefore, the number of linear arrangements of n different things is n × (number of circular arrangements of n different things). Hence, the number of circular arrangements of n different things is - n! 1/n × (number of linear arrangements of n different things) = = (n–1)! n Therefore note that : (i) The number of circular permutations of n different things taken all at a time is : (n – 1)!. If clockwise & anti-clockwise circular permutations are considered to be same, then it is : (n 1)! . 2 (ii) The number of circular permutations of n different things taking r at a time distinguishing clockwise & anticlockwise arrangements is : n Pr r Illustration 23 : In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls are together? (A) 5! × 5! (B) 5! × 4! (C) 1 5!2 (D) 1 5! × 4! 2 2 Solution : Leaving one seat vacant between two boys, 5 boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls sit in 5! ways. Hence the required number of ways = 4! × 5! Ans. (B) 10 E

JEE-Mathematics Illustration 24 : The number of ways in which 7 girls can stand in a circle so that they do not have same neighbours in any two arrangements ? (A) 720 (B) 380 (C) 360 (D) none of these Solution : Seven girls can stand in a circle by (7  1)! number of ways, because there is no difference in 2! anticlockwise and clockwise order of their standing in a circle. (7  1)!  = 360 Ans. (C) 2! Illustration 25 : The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is (A) 9! × 10! (B) 5(9!)2 (C) (9!)2 (D) none of these Solution : Ten pearls of one colour can be arranged in 1 .10  1! ways. The number of arrangements of 10 2 pearls of the other colour in 10 places between the pearls of the first colour = 10!  The required number of ways  1  9 ! 10 ! = 5 (9!)2 Ans. (B) 2 Illustration 26 : A person invites a group of 10 friends at dinner. They sit (i) 5 on one round table and 5 on other round table, (ii) 4 on one round table and 6 on other round table. Find the number of ways in each case in which he can arrange the guests. Solution : (i) The number of ways of selection of 5 friends for first table is 10C5. Remaining 5 friends are left for second table. The total number of permutations of 5 guests at a round table is 4!. Hence, the total number of 10!4!4! 10! arrangements is 10C5 × 4! × 4! =  5!5! 25 (ii) The number of ways of selection of 6 guests is 10C6. The number of ways of permutations of 6 guests on round table is 5!. The number of permutations of 4 guests on round table is 3! Therefore, total number of arrangements is : 10 C6 5 ! 3!  (1 0 ) ! 5!3!  (1 0 ) ! Ans. (B) 6!4! 24 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Do yourself-7 : ( i ) In how many ways can 3 men and 3 women be seated around a round table such that all men are always together ? ( i i ) Find the number of ways in which 10 different diamonds can be arranged to make a necklace. (i i i ) Find the number of ways in which 6 persons out of 5 men & 5 women can be seated at a round table such that 2 men are never together. ( i v ) In how many ways can 8 persons be seated on two round tables of capacity 5 & 3. 8. TOTAL NUMBER OF COMBINATIONS : ( a ) Given n different objects , the number of ways of selecting atleast one of them is, E nC1 + nC2 + nC3 +........+ nCn = 2n – 1. This can also be stated as the total number of combinations of n distinct things. ( b ) (i) Total number of ways in which it is possible to make a selection by taking some or all out of p + q + r +......things, where p are alike of one kind, q alike of a second kind, r alike of third kind & so on is given by : (p + 1) (q + 1) (r + 1).........–1. (ii) The total number of ways of selecting one or more things from p identical things of one kind, q identical things of second kind, r identical things of third kind and n different things is given by : (p + 1) (q + 1) (r + 1) 2n –1 11

JEE-Mathematics Illustration 27 : A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen. The number of ways of choosing P and Q so that P  Q =  is :- (A) 22n – 2nCn (B) 2n (C) 2n – 1 (D) 3n Solution : Let A = {a1, a2, a3, ..... an}. For ai  A, we have the following choices : (i) ai  P and ai Q (ii) ai  P and ai Q (iii) ai  P and ai Q (iv) ai  P and ai Q Out of these only (ii), (iii) and (iv) imply ai  P  Q. Therefore, the number of ways in which none of a1, a2, ....an belong to P  Q is 3n. Ans. (D) Illustration 28 : A student is allowed to select at most n books from a collection of (2n + 1) books. If the total number of ways in which he can select books is 63, find the value of n. Solution : Given student selects at most n books from a collection of (2n + 1) books. It means that he selects one book or two books or three books or ............ or n books. Hence, by the given condition- 2n+1C1 + 2n+1C2 + 2n+1C3 +.........+ 2n+1Cn = 63 ...(i) But we know that 2n+1C0 + 2n+1C1 + 2n+1C2 + 2n+1C3 + ....... + 2n+1C2n + 1 = 22n+1 ...(ii) Since 2n+1C0 = 2n+1C2n + 1 = 1, equation (ii) can also be written as 2 + (2n+1C1 + 2n+1C2 + 2n+1 C3 + ....... + 2n+1Cn) + (2n+1Cn+1 + 2n+1Cn+2 + 2n+1 Cn + 3 + ....... + 2n+1C2n–1 + 2n+1C2n) = 22n + 1  2 + (2n+1C1 + 2n+1C2 + 2n+1C3 + ......... + 2n+1Cn) + (2n+1Cn + 2n+1Cn–1 + ........ + 2n+1C2 + 2n+1C1) = 22n+1 ( 2n+1Cr = 2n+1C2n + 1 – r)  2 + 2 (2n+1C1 + 2n+1C2 + 2n+1C3 + ....... + 2n+1Cn) = 22n + 1 [from (i)]  2 + 2.63 = 22n+1  1+ 63 = 22n  64 = 22n  26 = 22n  2n = 6 Hence, n = 3. Ans. Illustration 29 : There are 3 books of mathematics, 4 of science and 5 of english. How many different collections can be made such that each collection consists of- (i) one book of each subject ? NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 (ii) at least one book of each subject ? Solution : (iii) at least one book of english ? (i) 3C1 × 4C1 × 5C1 = 60 Ans. (ii) (23–1) (24 – 1) (25 –1) = 7 × 15 × 31 = 3255 (iii) (25 – 1) (23) (24) = 31 × 128= 3968 Illustration 30 : Find the number of groups that can be made from 5 red balls, 3 green balls and 4 black balls, if at least one ball of all colours is always to be included. Given that all balls are identical except colours. Solution : After selecting one ball of each colour, we have to find total number of combinations that can be made from 4 red. 2 green and 3 black balls. These will be (4 + 1) (2 + 1) (3 + 1 ) = 60 Ans. Do yourself-8 : ( i ) There are p copies each of n different books. Find the number of ways in which atleast one book can be selected ? ( i i ) There are 10 questions in an examination. In how many ways can a candidate answer the questions, if he attempts atleast one question. 12 E

JEE-Mathematics 9. DIVISORS : Let N = pa. qb. rc ....... where p, q, r........ are distinct primes & a, b, c....... are natural numbers then : ( a ) The total numbers of divisors of N including 1 & N is = (a + 1) (b + 1) (c + 1)....... ( b ) The sum of these divisors is = (p0 + p1 + p2 + ....+ pa) (q0 + q1 + q2 + ....+ qb) (r0 + r1 + r2 + ....+ rc)... ( c ) Number of ways in which N can be resolved as a product of two factor is = 1 (a  1) (b  1) (c  1)...... if N is not a perfect square 2 1 (a  1) (b  1) (c  1)......  1 if N is a perfect square 2 ( d ) Number of ways in which a composite number N can be resolved into two factors which are relatively prime (or coprime) to each other is equal to 2n–1 where n is the number of different prime factors in N. Note : (i) Every natural number except 1 has atleast 2 divisors. If it has exactly two divisors then it is called a prime. System of prime numbers begin with 2. All primes except 2 are odd. (ii) A number having more than 2 divisors is called composite. 2 is the only even number which is not composite. (iii) Two natural numbers are said to be relatively prime or coprime if their HCF is one. For two natural numbers to be relatively prime, it is not necessary that one or both should be prime. It is possible that they both are composite but still coprime, eg. 4 and 25. (iv) 1 is neither prime nor composite however it is co-prime with every other natural number. (v) Two prime numbers are said to be twin prime numbers if their non-negative difference is 2 (e.g.5 & 7, 19 & 17 etc). (vi) All divisors except 1 and the number itself are called proper divisors. Illustration 31: Find the number of proper divisors of the number 38808. Also find the sum of these divisors. Solution : ( i ) The number 38808 = 23 . 32 . 72 . 11 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Hence the total number of divisors (excluding 1 and itself i.e.38808) Ans. = (3 + 1) (2 + 1) (2 + 1) (1 + 1) – 2 = 70 ( i i ) The sum of these divisors =(20 + 21 + 22 + 23) (30 + 31 + 32) (70 + 71 + 72) (110 + 111) – 1 – 38808 = (15) (13) (57) (12) – 1 – 38808 = 133380 – 1 – 38808 = 94571. Illustration 32: In how many ways the number 18900 can be split in two factors which are relative prime (or coprime) ? Solution: Here N = 18900 = 22 . 33 . 52 . 71 Number of different prime factors in 18900 = n = 4 Hence number of ways in which 18900 can be resolved into two factors which are relative prime (or coprime) = 24–1 = 23 = 8. Ans. Illustration 33: Find the total number of proper factors of the number 35700. Also find (i) sum of all these factors, (ii) sum of the odd proper divisors, (iii) the number of proper divisors divisible by 10 and the sum of these divisors. E 13

JEE-Mathematics Solution: 35700 = 52 × 22 × 31 × 71 × 171 The total number of factors is equal to the total number of selections from (5,5), (2,2), (3), (7) and (17), which is given by 3 × 3 × 2 × 2 × 2 = 72. These include 1 and 35700. Therefore, the number of proper divisors (excluding 1 and 35700) is 72 – 2 = 70 (i) Sum of all these factors (proper) is : (5° + 51 + 52) (2° + 21 + 22) (3° + 31) (7° + 71) (17° + 171) –1 –35700 = 31 × 7 × 4 × 8 × 18 – 1 – 35700 = 89291 (ii) The sum of odd proper divisors is : (5° + 51 + 52) (3° + 31) (7° + 71) (17° + 171) – 1 = 31 × 4 × 8 × 18 – 1 = 17856 – 1 = 17855 (iii) The number of proper divisors divisible by 10 is equal to number of selections from (5,5), (2,2), (3), (7), (17) consisting of at least one 5 and at least one 2 and 35700 is to be excluded and is given by 2 × 2 × 2 × 2 × 2 – 1= 31. Sum of these divisors is : (51 + 52) (21 + 22) (3° + 31) (7° + 71) (17° + 171) – 35700 = 30 × 6 × 4 × 8 × 18 – 35700 = 67980 Ans. Do yourself-9 : ( i ) Find the number of ways in which the number 94864 can be resolved as a product of two factors. ( i i ) Find the number of different sets of solution of xy = 1440. 1 0 . TOTAL DISTRIBUTION : ( a ) Distribution of distinct objects : Number of ways in which n distinct things can be distributed to p persons if there is no restriction to the number of things received by them is given by : pn ( b ) Distribution of alike objects : Number of ways to distribute n alike things among p persons so that each may get none, one or more thing(s) is given by n+p–1Cp–1. Illustration 34: In how many ways can 5 different mangoes, 4 different oranges & 3 different apples be distributed Solution : among 3 children such that each gets alteast one mango ? 5 different mangoes can be distributed by following ways among 3 children such that each gets atleast 1 : 311 221 Total number of ways :  3 5! !2 !  2 5 ! !   3 !  !1 !1 !2 !2  NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = 37 (as each fruit has 3 options).  Total number of ways   5!  5!   3 ! 3 7 Ans.  3!2! (2 !)3    Illustration 35: In how many ways can 12 identical apples be distributed among four children if each gets atleast Solution : 1 apple and not more than 4 apples. Let x,y,z & w be the number of apples given to the children.  x + y + z + w = 12 Giving one-one apple to each Now, x + y + z + w = 8 .......(i) Here, 0  x  3, 0  y  3, 0  z  3, 0  w  3 x = 3 – t1, y = 3 – t2, z = 3 – t3, w = 3 – t4. Putting value of x, y, z, w in equation (i) Put 12 – 8 = t1 + t2 + t3 + t4  t1 + t2 + t3 + t4 = 4 (Here max. value that t1, t2, t3 & t4 can attain is 3, so we have to remove those cases when any of ti getting value 4) = 7C3 – (all cases when atleast one is 4) = 7C3 – 4 = 35 – 4 = 31 Ans. 14 E

JEE-Mathematics Illustration 36: Find the number of non negative integral solutions of the inequation x + y + z  20. Solution : Illustration Let w be any number (0 < w < 20), then we can write the equation as : Solution : x + y + z + w = 20 (here x, y, z, w  0) Illustration Total ways = 23C3 Ans. 37: Find the number of integral solutions of x + y + z + w < 25, where x > – 2, y > 1, z  2, w  0. Given x + y + z + w < 25 x + y + z + w + v = 25 ........(i) Let x = –1 + t1, y = 2 + t2, z = 2 + t3, w = t4, v = 1 + t5 where (t1, t2, t3, t4  0) Putting value of x, y, z, w, v in equation (i)  t1 + t2 + t3 + t4 + t5 = 21. Ans. Number of solutions = 25C4 38: Find the number of positive integral solutions of the inequation x + y + z 150, where 0 < x  60, 0 < y  60, 0 < z  60. Solution : Let x = 60 – t1, y = 60 – t2, z = 60 – t3 (where 0  t1  59, 0  t2  59, 0  t3  59) Given x + y + z  150 or x + y + z – w = 150 (where 0 w  147) .......(i) Putting values of x, y, z in equation (i) 60 – t1 + 60 – t2 + 60 – t3 – w = 150 30 = t1 + t2 + t3 + w Total solutions = 33C3 Ans. 39: Find the number of positive integral solutions of xy = 12 Illustration Solution : xy = 12 xy = 22 × 31 (i) 3 has 2 ways either 3 can go to x or y (ii) 22 can be distributed between x & y as distributing 2 identical things between 2 persons NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 (where each person can get 0, 1 or 2 things). Let two person be 1 & 2  1 + 2 = 2  2+1C1 = 3C1 = 3 So total ways = 2 × 3 = 6. Alternatively : xy = 12 = 22 × 31 x  2 3a1 a2 0  a1  2 0  a2  1 y  2 3b1 b2 0  b1  2 0  b2  1 2 3  2 3a1 b1 a2 b221  a1 + b1 = 2  3C1 ways Ans. a2 + b2 = 1  2C1 ways Number of solutions = 3C1 × 2C1 = 3 × 2 = 6 E 15

JEE-Mathematics Illustration 40 : Find the number of solutions of the equation xyz = 360 when (i) x,y,z  N (ii) x,y,z  I Solution : ( i ) xyz = 360 = 23 × 32 × 5 (x,y,z  N) x = 2a1 3a2 5 a3 (where 0  a1  3, 0  a2  2, 0  a3  1) y = 2 b1 3b2 5 b3 (where 0  b1  3, 0  b2  2, 0  b3  1) z = 2c1 3c2 5 c3 (where 0  c1  3, 0  c2  2, 0  c3  1)  2 a1 3 a2 5 a3 .2 b1 3 b2 5 b3 .2 c1 3c2 5 c3  23  32  51  2 .3 .5  2  3  5a1 b1 c1 a2 b2 c2 a3 b3 c3 331  a1 + b1 + c1 = 3  5C2  10 a2 + b2 + c2 = 2 4C2  6 a3 + b3 + c3 = 1  3C2   3 Total solutions = 10 × 6 × 3 = 180. ( i i ) If x,y,z  I then, (a) all positive (b) 1 positive and 2 negative. Ans. Total number of ways = 180 + 3C2 × 180 = 720 Do yourself -10 : ( i ) In how many ways can 12 identical apples be distributed among 4 boys. (a) If each boy receives any number of apples. (b) If each boy receives atleast 2 apples. (ii) Find the number of non-negative integral solutions of the equation x + y + z = 10. (iii) Find the number of integral solutions of x + y + z = 20, if x  – 4, y  1, z  2 11. DEARRANGEMENT : There are n letters and n corresponding envelopes. The number of ways in which letters can be placed in the envelopes (one letter in each envelope) so that no letter is placed in correct envelope is n !   1  1  .....  (1)n  1 1! 2! n!    NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Proof : n letters are denoted by 1,2,3,........,n. Let Ai denote the set of distribution of letters in envelopes (one letter in each envelope) so that the ith letter is placed in the corresponding envelope. Then, n(Ai) = 1 × (n–1)! [since the remaining n–1 letters can be placed in n –1 envelops in (n–1)! ways] Then, n(Ai  Aj) represents the number of ways where letters i and j can be placed in their corresponding envelopes. Then, n(Ai  Aj) = 1 × 1 × (n–2)! Also n(Ai  Aj  Ak) = 1 × 1 × 1× (n–3)! Hence, the required number is n(A1'  A2'  ..... An') = n! – n(A1 A2......... An)     n !  n(A i )  n(A i  A j )  n(A i  A j  A k )  .......  (1)n n(A i  A 2 .....  A n ) = n! – [nC1(n–1)! – nC2(n–2)! + nC3(n–3)! + .......+ (–1)n–1 × nCn1]  n !  n! (n  1)! n! (n  2 ) ! ... .. ..  ( 1 ) n 1   n !   1  1  ........  (1)n  1 !(n  2!(n  2)!  1 1! 2! n!  1) !   16 E

JEE-Mathematics Illustration 41: A person writes letters to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that (i) all the letters are in the wrong envelopes. (ii) at least two of them are in the wrong envelopes. Solution : (i) The number of ways is which all letters be placed in wrong envelopes = 6! 1  1  1  1  1 – 1  1 = 720 1 – 1  1 1  1  1! 2! 3! 4! 5! 6 !  2 6 24  720  120 = 360 – 120 + 30 – 6 + 1 = 265. (i) The number of ways in which at least two of them in the wrong envelopes = 6C4 . 2! 1  1  1 + 6C3 . 3! 1  1  1  1 + 6C2 . 4! 1  1  1  1  1 1! 2 ! 1! 2! 3 ! 1! 2! 3! 4 ! + 6C1. 5! 1 1  1  1  1 – 1 + 6C0 6! 1  1  1  1  1 – 1  1 1! 2! 3! 4! 5 ! 1! 2! 3! 4! 5! 6 ! = 15 + 40 + 135 + 264 + 265 = 719. Ans. Do yourself - 11 : ( i ) There are four balls of different colours and four boxes of colours same as those of the balls. Find the number of ways in which the balls, one in each box, could be placed in such a way that a ball does not go to box of its own colour. Miscellaneous Illustrations : Illustration 42: In how many ways can a person go from point A to point B if he can travel only to the right or upward along the lines (Grid Problem) ? B(3,3) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 A(0,0) Solution : To reach the point B from point A, a person has to travel along 3 horizontal and 3 vertical strips. Therefore, we have to arrange 3H and 3V in a row. Total number of ways = 6 !  20 ways Ans. 3!3! Illustration 43: Find sum of all numbers formed using the digits 2,4,6,8 taken all at a time and no digit being repeated. Solution : All possible numbers = 4! = 24 If 2 occupies the unit's place then total numbers = 6 Hence, 2 comes at unit's place 6 times. Sum of all the digits occuring at unit's place = 6 × (2 + 4 + 6 + 8) Same summation will occur for ten's, hundred's & thousand's place. Hence required sum = 6 × (2 + 4 + 6 + 8) × (1 + 10 + 100 + 1000) = 133320 Ans. E 17

JEE-Mathematics Illustration 44: Find the sum of all the numbers greater than 1000 using the digits 0,1,2,2. Solution : (i) When 1 is at thousand's place, total numbers formed will be = 3!  3 2! (ii) When 2 is at thousand's place, total numbers formed will be = 3! = 6 (iii) When 1 is at hundred's, ten's or unit's place then total numbers formed will be- Thousand's place is fixed i.e. only the digit 2 will come here, remaining two places can be filled in 2! ways. So total numbers = 2! (iv) When 2 is at hundred's, ten's or unit's place then total numbers formed will be- Thousand's place has 2 options and other two places can be filled in 2 ways. So total numbers = 2 × 2 = 4 Sum = 103 (1 × 3 + 2 × 6) + 102 (1 × 2 + 2 × 4) + 101(1 × 2 + 2 × 4) + (1 × 2 + 2 × 4) = 15 × 103 + 103 + 102 + 10 = 16110 Ans. Ans. Illustration 45 : Find the number of positive integral solutions of x + y + z = 20, if x  y  z. Solution : x1 y = x + t1 t1  1 z = y + t2 t2  1 x + x + t1 + x + t1 + t2 = 20 3x + 2t1 + t2 = 20 (i) x = 1 2t1 + t2 = 17 t1 = 1,2 ......... 8  8 ways (ii) x = 2 2t1 + t2 = 14 t1 = 1,2 ......... 6  6 ways (iii) x = 3 2t1 + t2 = 11 t1 = 1,2 ......... 5  5 ways (vi) x = 4 2t1 + t2 = 8 t1 = 1,2,3  3 ways (v) x = 5 2t1 + t2 = 5 t1 = 1, 2  2 ways NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Total = 8 + 6 + 5 + 3 + 2 = 24 But each solution can be arranged by 3! ways. So total solutions = 24 × 3! = 144. Illustration 46: A regular polygon of 15 sides is constructed. In how many ways can a triangle be formed using the Solution : vertices of the polygon such that no side of triangle is same as that of polygon ? Select one point out of 15 point, therefore total number of ways = 15C1 Suppose we select point P1. Now we have to choose 2 more point which are not consecutive. since we can not select P2 & P15. Total points left are 12. Now we have to select 2 points out of 12 points P15 P1 P2 P3 which are not consecutive Total ways = 12–2 +1C2 = 11C2 P8 P6 Every select triangle will be repeated 3 times. P7 So total number of ways = 15 C1  11C2  275 3 18 E

JEE-Mathematics Alternative : First of all let us cut the polygon between points P1 & P15. Now there are 15 points on a straight line and we have to select 3 points out of these, such that the selected points are not consecutive. xOyOzOw Here bubbles represents the selected points, x represents the number of points before first selected point, y represents the number of points between Ist & IInd selected point, z represents the number of points between IInd & IIIrd selected point P15 P1 P2 and w represents the number of points after IIIrd selected point. P3 x + y + z + w = 15 – 3 = 12 here x > 0, y > 1, z > 1, w > 0 P8 P6 Put y = 1 + y' & z = 1 + z' (y' > 0, z' > 0) P7  x + y' + z' + w = 10 Total number of ways = 13C3 These selections include the cases when both the points P1 & P15 are selected. We have to remove those cases. Here a represents number of points between P1 & 3rd selected point & b represents number of points between 3rd selected point and P15  a + b = 15 – 3 = 12 (a > 1,b > 1) put a = 1 + t1 & b = 1 + t2 t1 + t2 = 10 Ans. Total number of ways = 11C1 = 11 Therefore required number of ways = 13C3 – 11C1 = 286 – 11 = 275 Illustration 47: Find the number of ways in which three numbers can be selected from the set {51, 52, 53,.....511} Solution : so that they form a G.P. Any three selected numbers which are in G.P. have their powers in A.P. Set of powers is = {1,2,.........6,7,.....11} By selecting any two numbers from {1,3,5,7,9,11}, the middle number is automatically fixed. Total number of ways = 6C2 Now select any two numbers from {2,4,6,8,10} and again middle number is automatically fixed. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Total number of ways = 5C2 Ans.  Total number of ways are = 6C2 + 5C2 = 15 + 10 = 25 ANSWERS FOR DO YOURSELF 1 : (i) 7 (ii) 3 2 : (i) r=4 3 : (i) 0 (ii) 450 (iii) 50C4 (iv) 20 (v) 120, 48 360 (iii) 840, 40 4 : (i) 10 (ii) 60 (iii) nC2.n! 5 : (i) 16 ! (ii) 6 : (i) 8! (2 !)8 8 ! 5n – 4n – 4n + 3n 60, 6th (ii) 7 : (i) 36 9! (iii) 5400 (iv) 2688 (ii)  181440 (iii) 23C2 8 : (i) (p + 1)n – 1 9 : (i) 23 2 10 : (i) (a) 15C3 (b) 7C3 ( i i ) 210 – 1 11 : (i) 9 (ii) 36 (i i ) 12C2 E 19

JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The total number of words which can be formed using all the letters of the word \"AKSHI\" if each word begins with vowel or terminates with vowel - (A) 84 (B) 12 (C) 48 (D) 60 2 . The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is - (A) 672 (B) 640 (C) 512 (D) none of these 3 . Out of seven consonants and four vowels, the number of words of six letters, formed by taking four consonants and two vowels is (Assume that each ordered group of letter is a word) - (A) 210 (B) 462 (C) 151200 (D) 332640 4 . A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2,3,4 & 5 without repetition. The total number of ways this can be done is - (A) 3125 (B) 600 (C) 240 (D) 216 5 . The number of ways in which 5 different books can be distributed among 10 people if each person can get at most one book is - (A) 252 (B) 105 (C) 510 (D) 10C . 5! 5 6 . Number of ways in which 9 different prizes can be given to 5 students, if one particular student receives 4 prizes and the rest of the students can get any numbers of prizes is - (A) 9C . 210 (B) 9C . 54 (C) 4 . 45 (D) none of these 4 5 7 . Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they are necessarily to be filled with either a red or a blue ball such that no two adjacent boxes can be filled with blue balls. How many different arrangements are possible, given that the balls of a given colour are exactly identical in all respects ? (A) 8 (B) 10 (C) 13 (D) 22 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 8 . Ten different letters of alphabet are given. Words with four letters are formed from these letters, then the number of words which have at least one letter repeated is - (A) 104 (B) 10P4 (C) 10C4 (D) 4960 9 . If all the letters of the word “QUEUE” are arranged in all possible manner as they are in a dictionary, then the rank of the word QUEUE is - (A) 15th (B) 16th (C) 17th (D) 18th 1 0 . Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is - (5 !)2 9! 9! (D) none of these (A) (B) 2 (C) 3 !(2 !)3 8 1 1 . The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if no two 'C's are together : 12! 13! 14! 13! (A) 13C3 . 5 !3 !2 ! (B) (C) (D) 11. 5!3!3!2! 5!3!2! 6! 20 E

JEE-Mathematics 1 2 . Number of numbers greater than a million and divisible by 5 which can be formed by using only the digits 1, 2, 1, 2, 0, 5 & 2 is - (A) 120 (B) 110 (C) 90 (D) none of these 1 3 . A set contains (2n + 1) elements. The number of subset of the set which contain at most n elements is : - (A) 2n (B) 2n+1 (C) 2n – 1 (D) 22n 1 4 . The maximum number of different permutations of 4 letters of the word “EARTHQUAKE” is - (A) 2910 (B) 2550 (C) 2190 (D) 2091 1 5 . The number of ways in which we can arrange n ladies & n gentlemen at a round table so that 2 ladies or 2 gentlemen may not sit next to one another is - (A) (n – 1)! (n – 2)! (B) (n)! (n – !)! (C) (n + 1)! (n)! (D) none of these 1 6 . The number of proper divisors of apbqcrds where a, b, c, d are primes & p, q, r, s  N is - (A) pqrs (B) (p + 1) (q + 1) (r + 1) (s + 1) – 4 (C) pqrs – 2 (D) (p + 1) (q + 1) (r + 1) (s + 1) – 2 1 7 . The sum of all numbers greater than 1000 formed by using the digits 1, 3, 5, 7 such that no digit is being repeated in any number is - (A) 72215 (B) 83911 (C) 106656 (D) 114712 1 8 . The number of way in which 10 identical apples can be distributed among 6 children so that each child receives atleast one apple is - (A) 126 (B) 252 (C) 378 (D) none of these 1 9 . Number of ways in which 25 identical pens can be distributed among Keshav, Madhav, Mukund and Radhika such that at least 1, 2, 3 and 4 pens are given to Keshav, Madhav, Mukund and Radhika respectively, is - (A) 18C4 (B) 28C3 (C) 24C3 (D) 18C3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 2 0 . There are (p + q) different books on different topics in Mathematics. (p  q) If L = the number of ways in which these books are distributed between two students X and Y such that X get p books and Y gets q books. M = The number of ways in which these books are distributed between two students X and Y such that one of them gets p books and another gets q books. N = The number of ways in which these books are divided into two groups of p books and q books then - (A) L = N (B) L = 2M = 2N (C) 2L = M (D) L = M 2 1 . Number of dissimilar terms in the expansion of (x1 + x2 + ...... + xn)3 is - n2 (n  1)2 n(n  1)(n  2) (C) n+1C2 + n+1C3 n3  3n2 (A) (D) (B) 6 4 4 2 2 . A persons wants to invite one or more of his friend for a dinner party. In how many ways can he do so if he has eight friends : - (A) 28 (B) 28 – 1 (C) 82 (D) 8C1 + 8C2 + .....+ 8C8 E 21

JEE-Mathematics 2 3 . If P(n, n) denotes the number of permutations of n different things taken all at a time then P(n, n) is also identical to :- (A) n.P(n – 1, n – 1) (B) P(n, n – 1) (C) r! . P(n, n – r) (D) (n – r) . P(n, r) where 0  r  n 2 4 . Which of the following statement(s) is/are true :- (A) 100C50 is not divisible by 10 (B) n(n – 1)(n – 2) .........(n – r + 1) is always divisible by r! (n  N and 0  r  n) (C) Morse telegraph has 5 arms and each arm moves on 6 different positions including the position of rest. Number of different signals that can be transmitted is 56 – 1. (D) There are 5 different books each having 5 copies. Number of different selections is 65 –1. BRAIN TEASERS ANSWER KEY EXERCISE-2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. A A C D D A C D C C Que. 11 12 13 14 15 16 17 18 19 20 Ans. A B D C B D C A D A,C Que. 21 22 23 24 Ans. B,C B,D A,B,C A,B,D 22 E

EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . 5 Indian & 5 American couples meet at a party & shake hands. If no wife shakes hands with her own husband & no Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party is - (A) 95 (B) 110 (C) 135 (D) 150 2 . The number of ways in which a mixed double tennis game can be arranged from amongst 9 married couple if no husband & wife plays in the same game is - (A) 756 (B) 3024 (C) 1512 (D) 6048 3 . There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of \"n red balls but not necessarily all the green balls\" is xC then - y (A) x = m + n, y = m (B) x = m + n + 1, y = m (C) x = m + n + 1, y = m + 1 (D) x = m + n, y = n 4 . Number of different words that can be formed using all the letters of the word “DEEPMALA” if two vowels are together and the other two are also together but separated from the first two is - (A) 960 (B) 1200 (C) 2160 (D) 1440 5 . In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches. The number of ways in which the series can be won by India, if no match ends in a draw is - (A) 126 (B) 252 (C) 225 (D) none of these 6 . A road network as shown in the figure connect four cities. In how many ways can you start from any city (say A) and come back to it without travelling on the same road more than once ? (A) 8 (B) 12 (C) 9 (D) 16 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 7 . The number of ways of choosing a committee of 2 women & 3 men from 5 women & 6 men, if Mr. A refuses to serve on the committee if Mr. B is a member & Mr. B can only serve, if Miss C is the member of the committee, is - (A) 60 (B) 84 (C) 124 (D) none of these 8 . Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be done if A must have either B or C on his right and B must have either C or D on his right is - (A) 36 (B) 12 (C) 24 (D) 18 9 . Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is - (A) 22222200 (B) 11111100 (C) 55555500 (D) 20333280 1 0 . N = 22. 33.54.7, then - (A) Number of proper divisors of N(excluding 1 & N) is 118 (B) Number of proper divisors of N(excluding 1 & N) is 120 (C) Number of positive integral solutions of xy = N is 60 (D) Number of positive integral solutions of xy = N is 120 E 23

JEE-Mathematics 1 1 . Sameer has to make a telephone call to his friend Harish, Unfortunately he does not remember the 7 digit phone number. But he remembers that the first three digits are 635 or 674, the number is odd and there is exactly one 9 in the number. The maximum number of trials that Sameer has to make to be successful is - (A) 10,000 (B) 3402 (C) 3200 (D) 5000 1 2 . Let P denotes the number of ways in which three people can be selected out of ‘n’ people sitting in a row, if no n two of them are consecutive. If P – P = 15 then the value of ‘n’ is - n+1 n (A) 7 (B) 8 (C) 9 (D) 10 1 3 . The number of solutions of x1 + x2 + x3 = 51 (x1, x2, x3 being odd natural numbers) is : - (A) 300 (B) 325 (C) 330 (D) 350 1 4 . The number of positive integral solutions of the equation x1x2x3 = 60 is : - (A) 54 (B) 27 (C) 81 (D) none of these 1 5 . Total number of even divisors of 2079000 which are divisible by 15 are - (A) 54 (B) 128 (C) 108 (D) 72 1 6 . The number of five digit numbers that can be formed using all the digits 0, 1, 3, 6, 8 which are - (A) divisible by 4 is 30 (B) greater than 30,000 and divisible by 11 is 12 (C) smaller than 60,000 when digit 8 always appears at ten's place is 6 (D) between 30,000 and 60,000 and divisible by 6 is 18. 1 7 . All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once and not divisible by 5 are arranged in the increasing order. Then - (A) 1800th number in the list is 3124567 (B) 1897th number in the list is 4213567 (C) 1994th number in the list is 4312567 (D) 2001th number in the list is 4315726 BRAIN TEASERS ANSWER KEY EXERCISE-2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Que. 1 2 3 4 5 6 7 8 9 10 A A,D Ans. C C B D A B C D E Que. 11 12 13 14 15 16 17 Ans. B B B A C A,B,D B,D 24

JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . 5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. Number of ways in which the balls can be placed so that no box remains empty, if : Column-I Column-II (A) balls are identical but boxes are different (p) 2 (B) balls are different but boxes are identical (q) 2 5 (C) balls as well as boxes are identical (r) 5 0 (D) balls as well as boxes are identical but boxes are kept in a row (s) 6 2 . Consider all the different words that can be formed using the letters of the word HAVANA, taken 4 at a time. Column-I Column-II (A) Number of such words in which all the 4 letters are different (p) 36 (B) Number of such words in which there are 2 alike letters & (q) 4 2 2 different letters. (r) 3 7 (C) Number of such words in which A's never appear together (s) 2 4 (D) If all such 4 letters words are written, by the rule of dictionary then the rank of the word HANA 3 . Column-I Column-II (A) 24C2 + 23C2 + 22C2 + 21C2 + 20C2 + 20C3 is equal to (p) 102 (4,4) (q) 2300 (B) In the adjoining figure number of progressive 4 (2,2) ways to reach from (0,0) to (4, 4) passing 3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 through point (2, 2) are 2 (particle can move on horizontal or vertical line) 1 0 (r) 8 2 1 2 34 (C) The number of 4 digit numbers that can be made with the digits 1, 2, 3, 4, 3, 2 (D) If 500 !   0, then the maximum natural value of k is equal to (s) 3 6    14 k  (where {.} is fractional part function) ASSERTION & REASON These questions contain, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1 . Statement-I : If a polygon has 45 diagonals, then its number of sides is 10. Because Statement-II : Number of ways of selecting 2 points from n non collinear points is nC2. (A) A (B) B (C) C (D) D E 25

JEE-Mathematics 2 . Statement-I : The expression n!(20 – n)! is minimum where n = 10 Because Statement-II : 2pCr is maximum where r = p, where p is a constant. (A) A (B) B (C) C (D) D 3 . Statement-I : The number of non negative integral solutions of x1 + x2 + x3 + ...... + xn = r is r+n–1Cr. Because Statement-II : The number of ways in which n identical things can be distributed among r students is n+r–1Cn. (B) B (C) C (D) D (A) A 4 . Statement-I : If a, b, c are positive integers such that a + b + c  8, then the number of possible values of the ordered triplets (a, b, c) is 56. Because Statement-II : The number of ways in which n distinct things can be distributed among r girls such that each get at least one is n–1C . r–1 (A) A (B) B (C) C (D) D 5 . Statement-I : Number of terms in the expansion of (x1 + x2 + x3 + .... + x11)6 = 16C6. Because Statement-II : Number of ways of distributing n identical things among r persons when each person get zero or more things = Cn+r–1 n (A) A (B) B (C) C (D) D 6 . Statement-I : Number of ways in which 400 different things can be distributed between Ramu & Shamu so that each receives 200 things > Number of ways in which 400 different things can be distributed between Sita & Geeta. So that Sita receives 238 things & Geeta receives 162 things. Because Statement-II : Number of ways in which (m + n) different things can be distributed between two receivers such that one receives m and other receives n is equal to m+nCm, for any two non-negative integers m and n. (A) A (B) B (C) C (D) D 7 . Statement-I : The number of positive integral solutions of the equation xyzw = 770 is 28. Because Statement-II : The number of ways of selection of atleast one thing from n things of which 'p' are alike of one NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 kind, q are alike of 2nd kind and rest of the things are different is (p + 1)(q + 1) 2n–(p+q) – 1. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 S = {0, 2, 4, 6, 8}. A natural number is said to be divisible by 2 if the digit at the unit place is an even number. The number is divisible by 5, if the number at the unit place is 0 or 5. If four numbers are selected from S and a four digit number ABCD is formed. On the basis of above information, answer the following questions : 1 . The number of such numbers which are even (all digits are different) is (A) 60 (B) 96 (C) 120 (D) 204 2 . The number of such numbers which are even (all digits are not different) is (A) 404 (B) 500 (C) 380 (D) none of these 3 . The number of such numbers which are divisible by two and five (all digits are not different) is (A) 125 (B) 76 (C) 65 (D) 100 26 E

JEE-Mathematics Comprehension # 2 Let p be a prime number and n be a positive integer, then exponent of p is n! is denoted by Ep (n!) and is given by n  n   n  n  E (n!) =  p  +  p2  +  p3  + ..... +  pk  p       where pk < n < pk+1 and [x] denotes the integral part of x. If we isolate the power of each prime contained in any number N, then N can be written as N = 21 · 32 · 53 · 74 .... where i are whole numbers. On the basis of above information, answer the following questions : 1 . The exponent of 7 in 100C is - 50 (A) 0 (B) 1 (C) 2 (D) 3 (D) 26 2 . The number of zeros at the end of 108! is - (D) none of these (A) 10 (B) 13 (C) 25 3 . The exponent of 12 in 100! is - (A) 32 (B) 48 (C) 97 Comprehension # 3 We have to choose 11 players for cricket team from 8 batsmen. 6 bowlers, 4 allrounders and 2 wicketkeeper, in the following conditions. On the basis of above information, answer the following questions : 1 . The number of selections when at most 1 allrounder and 1 wicketkeeper will play - (A) 4C1 .14C10 + 2C1 .14C10 + 4C1 .2C1 .14C9 + 14C11 (B) 4C1 . 15C11 + 15C11 (C) 4C1 . 15C10 + 15C11 (D) none of these 2 . Number of selections when 2 particular batsmen don't want to play, if a particular bowler will play - (A) 17C10 + 19C11 (B) 17C10 + 19C11 + 17C11 (C) 17C10 + 20C11 (D) 19C10 + 19C11 3 . Number of selections when a particular batsman and a particular wicketkeeper don't want to play together - (A) 2 18C10 (B) 19C11 + 18C10 (C) 19C10 + 19C11 (D) none of these NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  Match the Column 1 . (A)  (s), (B)  (q), (C)  (p), (D)  (s) 2 . (A)  (s), (B)  (p), (C)  (q), (D)  (r)  3 . (A)(q), (B)(s), (C)(p), (D)(r) 6. C 7. B  Assertion & Reason 1. D 2. A 3. A 4. C 5. A 3. B E Comprehension Based Questions 3. B Comprehension # 1 : 1. B 2. A 3. B Comprehension # 2 : 1. A 2. C Comprehension # 3 : 1. A 2. A 27

JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE EXERCISE - 04 [A] 1 . (a) Prove that : nPr = n–1Pr + r. n–1Pr–1 (b) If 20Cr+2 = 20C2r–3 find 12Cr (c) Find r if 15C3r = 15Cr+3 (d) Find the ratio 20Cr to 25Cr when each of them has the greatest value possible. 2 . In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be 3 out of ABC A'B'C', but never AA', BB' or CC' together ? 3 . How many 4 digit numbers are there which contain not more than 2 different digits ? 4 . An examination paper consists of 12 questions divided into parts A & B. Part-A contains 7 questions & Part -B contains 5 questions. A candidate is required to attempt 8 questions selecting atleast 3 from each part. In how many maximum ways can the candidate select the questions ? 5. The straight lines 1, 2, and  are parallel & lie in the same plane. A total of m points are taken on the line 1, 3 n points on  and k points on 3. How many maximum number of triangles are there whose vertices are at these 2 points ? 6 . Prove that if each of m points in one straight line be joined to each of n in another by straight lines terminated NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 1 by the points, then excluding the given points, the lines will intersect mn(m – 1) (n – 1) times. 4 7 . A man has 7 relatives, 4 of them are ladies & 3 gentlemen; his wife has also 7 relatives, 3 of them are ladies & 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies & 3 gentlemen so that there are 3 of the man's relatives & 3 of the wife's relatives ? 8 . 5 boys & 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are together & the other 2 are also together but separated from the first 2. 9 . In how many ways 8 persons can be seated on a round table (a) If two of them (say A and B) must not sit in adjacent seats ? (b) If 4 of the persons are men and 4 ladies and if no two men are to be in adjacent seats? (c) If 8 persons constitute 4 married couples and if no husband and wife, as well as no two men are to be in adjacent seats ? 1 0 . There are 2 women participating in a chess tournament. Every participant played 2 games with the other par- ticipants. The number of games that the men played between themselves exceeded by 66 as compared to the number of games that the men played with the women. Find the number of participants & the total number of games played in the tournament. 28 E

NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 JEE-Mathematics 1 1 . In how many ways can you divide a pack of 52 cards equally among 4 players. In how many ways the cards can be divided in 4 sets, 3 of them having 17 cards each & the 4th with 1 card ? 1 2 . (a) How many divisors are there of the number 21600 ? Also find the sum of these divisors. (b) In how many ways the number 7056 can be resolved as a product of 2 factors. (c) Find the number of ways in which the number 300300 can be split into 2 factors which are relatively prime. 1 3 . How many different ways can 15 candy bars be distributed between Ram, Shyam, Ghanshyam and Balram, if Ram can not have more than 5 candy bars and Shyam must have at least two ? Assume all candy bars to be alike. 1 4 . Find the sum of all numbers greater than 10000 formed by using the digits 0, 1, 2, 4, 5 & no digit being repeated in any number. 1 5 . Find the number of ways in which the letters of the word 'MUNMUN' can be arranged so that no two alike letters are together. 1 6 . A shop sells 6 different flavours of ice-creams. In how many ways can a customer choose 4 ice-cream cones if (a) they are all of different flavours ? (b) they are non necessarily of different flavours ? (c) they contain only 3 different flavours ? (d) they contain only 2 or 3 different flavours ? 1 7 . Find the number of ways in which a selection of 100 balls, can be made out of 100 identical red balls, 100 identical blue balls & 100 identical white balls. 1 8 . There are 5 balls of different colours & 5 boxes of colours same as those of the balls. The number of ways in which the balls, one in each box could be placed such that exactly one ball goes to the box of its own colour. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 143 2. 960 3. 576 4. 420 1 . (b) 792, (c) r = 3 (d) 485 8. 43200 9. (a) 5.(6!) (b) 3!4! (c) 12 4025 (b) 23 (c) 32 1 3 . 440 17. 5151 18. 45 5. Cm + n + k – (mC3 + nC3 + kC3) 7. 3 52! 52! 10 .13, 156 11. ; 1 2 . (a) 72 ; 78120 (13 !)4 3 !(17!)3 1 4 .3119976 15. 30 1 6 . (a) 15; (b) 126 ; (c) 60 ; (d) 105 E 29

JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE - 04 [B] 1 . There are counters available in 7 different colours. Counters are all alike except colour and they are atleast ten of each colour. Find the number of ways in which an arrangement of 10 counters can be made. How many of these will have counters of each colour ? 2 . How many integral solutions are there for the equation; x + y + z + w = 29 when x > 0, y >1, z>2&w0? 3 . A party of 10 consists of 2 Americans, 2 Britishmen, 2 Chinese & 4 men of other nationalities (all different). Find the number of ways in which they can stand in a row so that no two men of the same nationality are next to one another. Find also the number of ways in which they can sit at a round table. 4 . Find the number of words each consisting of 3 consonants & 3 vowels that can be formed from the letters of the word \"CIRCUMFERENCE\". In how many of these 'C's will be together. 5 . Find the number of ways in which the number 30 can be partitioned into three unequal parts, each part being a natural number. What this number would be if equal parts are also included. 6 . Prove by combinatorial argument that : (a) n+1Cr = nCr + nCr–1 (b) n+mCr = nC0 . mCr + nC1 . mCr–1 + nC2 . mCr–2 + ..... + nCr . mC0 7 . How many 6 digits odd numbers greater than 60,0000 can be formed from the digits 5,6,7,8,9.0 if (a) repetitions are not allowed ? (b) repetitions are allowed ? 8 . All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once and not divisible by 5 are arranged in the increasing order. Find the (2004)th number in this list. 9 . A firm of Chartered Accountants in Bombay has to send 10 clerks to 5 different companies, two clerks in each. Two of the companies are in Bombay and the others are outside. Two of the clerks prefer to work in Bombay while three others prefer to work outside. In how many ways can the assignment be made if the preferences are to be satisfied ? 1 0 . There are 5 white, 4 yellow, 3 green, 2 blue & 1 red ball. The balls are all identical except colour. These are to NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 be arranged in a line at 5 places. Find the number of distinct arrangements. 1 1 . A crew of an eight oar boat has to be chosen out of 11 men five of whom can row on stroke side only, four on the bow side only and the remaining two on either side. How many different selections can be made? 1 2 . There are n straight lines in a plane, no 2 of which are parallel & no 3 pass through the same point. Their points of intersection are joined. Show that the number of fresh lines introduced is n(n  1)(n  2)(n  3) . 8 1 3 . There are 20 books on Algebra & Calculus in our library. Prove that the greatest number of selections each of which consists of 5 books on each topic is possible only when there are 10 books on each topic in the library. 1 4 . Find the number of ways to invite one of the three friends for dinner on 6 successive nights such that no friend is invited more than 3 times. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1. 710 ;  49 10 ! 2. 2600 3. (47)8! ; (244)6! 4 . 22100, 52 5 . 61, 75  6 9. 5400 14. 510 10. 2111 11. 145 7 . (a) 240 (b) 15552 8. 4316527 30 E

EXERCISE - 05 [A] JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . Numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is allowed), are [AIEEE 2002] (1) 350 (2) 375 (3) 450 (4) 576 2 . A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is [AIEEE 2002] (1) 216 (2) 240 (3) 600 (4) 3125 3 . Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are [AIEEE 2002] (1) 192 (2) 375 (3) 400 (4) 720 4 . The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by [AIEEE 2003] (1) 6! × 5! (2) 30 (3) 5! × 4! (4) 7! × 5! 5 . A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five question. The number of choices available to him is [AIEEE 2003] (1) 140 (2) 196 (3) 280 (4) 346 6 . If nCr denots the number of combinations of n things taken r at a time, then the expression n Cr1  nCr1  2  nCr equals [AIEEE 2003] (1) n+2Cr (2) Cn+2 (3) n+1Cr (4) Cn+1 r+1 r+1 7 . How many ways are there to arrange the letters in the word 'GARDEN' with the vowels in alphabetical order? [AIEEE 2004] (1) 120 (2) 240 (3) 360 (4) 480 8 . The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [AIEEE 2004] (1) 5 (2) 21 (3) 38 (4) 8 C3 9 . If the letters of the word 'SACHIN' are arranged in all possible ways and these words are written out as in dictionary, then the word 'SACHIN' appears at serial number [AIEEE 2005] (1) 602 (2) 603 (3) 600 (4) 601 6 1 0 . The value of 50 C4  C56 r is [AIEEE 2005] 3 r 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 (1) 56C4 (2) 56C3 (3) 55C3 (4) 55C4 1 1 . At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidated and 4 are to be elected. If a voter votes for at least one candidates, then the number of ways in which he can vote is [AIEEE-2006] (1) 385 (2) 1110 (3) 5040 (4) 6210 1 2 . The set S = {1, 2, 3,.....,12} is to partitioned into three sets A, B, C of equal size. Thus,A B C =S, A  B = B  C = C A = , then number of ways to partition S are- [AIEEE-2007] 12! 12! 12! 12! (1) 3!(3!)4 (2) (4!)3 (3) (3!)4 (4) 3!(4!)3 1 3 . In a shop there are five types of ice–creams available. A child buys six ice–creams. [AIEEE 2008] Statement –1 : The number of different ways the child can buy the six ice–creams is 10C . 5 Statement –2 : The number of different ways the child can buy the six ice–creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. (1) Statement –1 is false, Statement –2 is true (2) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1 (3) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1 (4) Statement–1 is true, Statement–2 is false E 31

JEE-Mathematics 1 4 . From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is :- [AIEEE 2009] (1) At least 750 but less than 1000 (2) At least 1000 (3) Less than 500 (4) At least 500 but less than 750 1 5 . There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is :- [AIEEE-2010] (1) 3 (2) 36 (3) 66 (4) 108 1 6 . Statement - 1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3. [AIEEE-2011] Statement - 2 : The number of ways of choosing any 3 places from 9 different places is 9C3. (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is false, Statement-2 is true (3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. 1 7 . There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points, then : [AIEEE-2011] (1) N > 190 (2) N < 100 (3) 100 < N < 140 (4) 140 < N < 190 1 8 . Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is :- [AIEEE-2012] (1) 879 (2) 880 (3) 629 (4) 630 1 9 . Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is [JEE (Main)-2013] (1) 256 (2) 220 (3) 219 (4) 211 2 0 . Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 – Tn = 10, then the value of n is : [JEE (Main)-2013] (1) 7 (2) 5 (3) 10 (4) 8 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 2 1 1 1 2 Que. 16 17 18 19 20 2324112124 Ans 3 2 1 3 2 32 E

JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ? [JEE 2000, Screening 1M] (A) 16 (B) 36 (C) 60 (D) 180 2 . ( a ) Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of ‘n’ sides. If Tn+1 – Tn = 21, then ‘n’ equals - [JEE 2001, Screening 1+1M] (A) 5 (B) 7 (C) 6 (D) 4 ( b ) Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is - (A) 14 (B) 16 (C) 12 (D) 8 3 . The number of arrangements of the letters of the word BANANA in which two 'N's do not appear adjacently is - [JEE 2002, Screening 3M] (A) 40 (B) 60 (C) 80 (D) 100 4 . Number of points with integral co-ordinates that lie inside a triangle whose co-ordinates are (0, 0), (0, 21) and (21, 0) [JEE 2003, Screening 3M] (A) 210 (B) 190 (C) 220 (D) none of these (n2 )! 5 . Using permutation or otherwise prove that (n !)n is an integer, where n is a positive integer. [JEE 2004, (Mains) 2M out of 60] 6 . A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit length by drawing parallel lines as shown then number of rectangles possible with odd side lengths is - (A) (m + n + 1)2 [JEE 2005, Screening 3M] (C) m2 n2 (B) 4m + n – 1 (D) mn (m + 1) (n + 1) 7 . If r, s, t are the prime numbers and p, q are the positive integers such that the LCM of p & q is r2t4s2, then the number of ordered pair (p, q) is : [JEE 2006, (3M, –1M) out of 184] (A) 252 (B) 254 (C) 225 (D) 224 8 . The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is - [JEE 2007, 3M] (A) 360 (B) 192 (C) 96 (D) 48 9 . Consider all possible permutations of the letters of the word ENDEANOEL. [JEE 2008, 6M] Match the Statements / Expressions in Column I with the Statements / Expressions in Column II. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 Column I Column II (A) The number of permutations containing the word ENDEA is (p) 5! (B) The number of permutations in which the letter E occurs in the first (q) 2 × 5! and the last positions is (r) 7 × 5! (C) The number of permutations in which none of the letters D, L, N (s) 21 × 5! occurs in the last five positions is (D) The number of permutations in which the letters A, E, O occur only in odd positions is 1 0 . The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is - [JEE 2009, 3M, –1M] (A) 55 (B) 66 (C) 77 (D) 88 1 1 . Let S = {1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to - [JEE 10, 5M, –2M] (A) 25 (B) 34 (C) 42 (D) 41 1 2 . The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is - [JEE 2012, 3M, –1M] (A) 75 (B) 150 (C) 210 (D) 243 E 33

JEE-Mathematics Paragraph for Question 13 and 14 : Let an denotes the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b = the number of such n-digit integers ending with digit 1 and c = the number of such nn n-digit integers ending with digit 0. 1 3 . The value of b is (B) 8 (C) 9 [JEE 2012, 3M, –1M] 6 (D) 11 (A) 7 1 4 . Which of the following is correct ? [JEE 2012, 3M, –1M] (A) a = a + a (B) c  c + c (C) b  b + c (D) a = c + b 17 16 15 17 16 15 17 16 16 17 17 16 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#12\\ENG\\01.PERMUTATION-COMBINATION.p65 1. C 2 . (a) B; (b) 14 3. A 4. B 6. C 7. C 8. C 10. C 11. D 12. B 13. B 14. A 9 . (A)(p), (B)(s), (C)(q), (D)(q) 34 E