SLM_(1)

x+8 x 1 x+6 x 3 The expansion is (x + 8) (x – 3) – (x – 1) (x + 6) C x2 + 5x – 24 – (x2 + 5x – 6) D x2 + 5x – 24 – x2 – 5x + 6 E – 18 5.2 Simultaneous Equations Simultaneous Equations can be solved conveniently by using determinants. The procedure is explained in the solution of the following equations: CU IDOL SELF LEARNING MATERIAL (SLM)

96 Business Mathematics and Statistics a1x + b1y + c1 = 0 … (1) … (2) a2x + b2y + c2 = 0 …(3) Multiplying eqn. (1) by b2 and eqn (2) by b1, we have, … (4) a1b2x + b1b2y + c1b2 = 0 a2b1x + b1b2y + c2b1 = 0 Subtracting eqn (4) from eqn (3), we get (a1b2 – a2b1) x + c1b2 – c2b1 = 0 ? (a1b2 – a2b1) x = b1c2 – b2c1 b1c2 b2 c1 provided a1b2 – a2b1 0 x = a1 b2 a2 b1 c1a2 c2 a1 Similarly, y = a b a b 12 21 x y 1 Therefore, b1c2 b2 c1 c1a2 c2 a1 a1 b2 a2 b1 provided a1b2 – a2b1 0 Using determinants, we can write the above as x = y =1 b1 c1 c1 a1 a1 b1 c2 a2 a2 b2 bc 22 Example 3:Using determinants, solve 2x – 3y = 3

5x + 2y = 36 The given equation can be written as 2x – 3y – 3 = 0 5x + 2y – 36 = 0 ?x = y = 1 33 32 23 2 36 36 5 52 CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 97 x y1 == ? 15+72 4+15 108+6 x y 1 = = ? 57 19 114 ? x = 114/19 = 6 y = 57/19 = 3 5.2 The Linear Equations in Two Unknowns Given three equations, we proceed to find the condition that they are satisfied by the same values of x and y. Suppose the given equation are y … (1) l1x + m1y + n1 = 0 … (2) l2x + m2y + n2 = 0 … (3) l3x + m3y + n3 = 0 =1 Solving (2) and (3) we get, x= m2 n2 n2 l2 l2 m2 m n l3 m3 3 n3 3 l3 provided l2m3 – l3m2 0 ? x= m2 n2 and y = n2 l2

m3 n3 n3 l3 l2 m2 l2 m2 l3 m3 l3 m3 If these values of x and y are to satisfy equation (1), then the condition is: m n + n1 = 0 2 n2 2 l2 l1 + m1 n3 l3 m3 n3 CU IDOL SELF LEARNING MATERIAL (SLM)

98 Business Mathematics and Statistics m m l2 2 l2 2 l3 m3 l3 m3 n2 l2 lm i.e., l m2 n2 + m +n 2 2 =0 1 1n 1 l3 m3 m3 n3 3 l3 l2 n2 or l m2 n2 – m + n l2 m2 = 0 1 1 1 l3 m3 m3 n3 l3 n3 n2 l2 = n2l3– n3l2 = – (n3l2 – n2l3) because n3 l3 3 – (e) 2 n2 l3 n3 Thus the condition is l1 (m2n3 – m3n2) – m1 (l2n3 – n2l3) + n1 (l2m3 – l3m2) = 0 The left hand side of the above expression is written in the form

l1 m1 n1 l2 m2 n2 l3 m3 n3 Which is called a determinant of the third order. The expression l1 (m2n3 – m3n2) – m1 (l2n3 – n2l3) + n1 (l2m3 – l3m2) is called the expansion of the above determinant. It may be noted that a determinant of the third order has to be expanded in terms of the determinants of the second order. Example 4: Expand lab amc bc n CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 99 The expansion is mc ac am l c n –a b n +b b c 4. 1 (mn – c2) – a (an – bc) + b (ac – bm) 5. lmn – lc2 – a2n + abc + abc – b2m 6. lmn + 2 abc – a2n – b2m – c2l Example 5: Evaluate: 345 5 34 453 The determinant 3 4 5 4 53 =3 5 3 –4 4 3 +5 4 5 z 3(9–20)–4(15–16)+5(25–12) z 3(–11)–4(–1)+5(13) z –33+4+65 z 36 Example 6: Are the following equations consistent?

x + 2y + 3 = 0 3x + 5y + 7 = 0 7x – 4y – 15 = 0 The above equations would be consistent only if the determinanat. 12 3 35 7 74 is zero 15 CU IDOL SELF LEARNING MATERIAL (SLM)

100 Business Mathematics and Statistics Verifying, we use 1(–75+28)–2(–45–49)+3(–12–35) i.e., – 75 + 28 + 188 – 141 i.e., 216 – 216 =0 Hence verified. The given equations are consistent. Example 7: Solve 1x 7 25 4 =0 x 7 15 Expanding the determinants, we have, 1 (– 75 + 28) – x (– 30 + 4x) + 7 (14 – 5x) = 0 i.e., – 47 + 30 – 4x2 + 98 – 35x = 0 i.e., – 4x2 – 5x + 51 = 0 i.e., 4x2 + 5x – 51 = 0 i.e., 4x2 – 12x + 17x – 51 = 0 i.e., 4x (x – 3) + 17 (x – 3) = 0 i.e., (4x + 17) (x – 3) = 0

? x = – 17/4 and x = 3 5.4 Theorems on Determinants Theorem 1 : The value of a determinant is not altered if the rows and columns are interchanged. Consider the determinant x1 x2 x3 D= yy y z1 z2 12 3 z3 CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 101 Let the determinant that is obtained by interchanging rows and columns be denoted by D', that is, x1 y1 z1 xyz D' = 2 2 2 6. 3 y3 z3 We shall prove that D’ = D. Expanding the determinant D', we get, D' = x1 (y2z3 – y3z2) – y1 (x2z3 – x3z2) + z1 (x2y3 – x3y2) = x1 (y2z3 – y3z2) – x2 (y1z3 – y3z1) + x3 (y1z2 – y2z1) =D ? D'= D Hence the theorem is proved. Theorem 2: If two rows or two columns of a determinant be interchanged then the determinant is altered in sign only. x1 x2 x3 yy y Suppose, D = 1 2 3 z1 z2 z3

Let the determinant that is obtained by interchanging first and second rows of D be denoted by D', then y 1 y2 y3 D' º x1 x2 x3 z1 z2 z3 To show that D' = – D, Now D' = y1 (x2z3 – x3z2) – y2 (x1z3 – x3z1) + y3 (x1z2 – x2z1) = { x1 (y2z3 – y3z2) – x2 (y1z3 – y3z1) + x3 (y1z2 – y2z1) } = –D Hence the theorem is proved. CU IDOL SELF LEARNING MATERIAL (SLM)

102 Business Mathematics and Statistics Theorem 3: If in a determinant, two rows (or two columns) are identical then the value of the determinant is zero. Suppose D denotes the determinant which has two rows identical Let D' denote the determinant obtained by interchanging the two identical rows. Then, by virtue of theorem 2, it follows that D' = – D … (1) In fact D' is the same as D, as the two rows that are interchanged are identical. Thus, D'= D … (2) From results (1) and (2) it follows that D= –D = 2D= 0 ?D= 0 Hence the theorem is proved. Theorem 4:If all the elements of one row (or column) of a determinant are multiplied by a member ‘p’ then the value of the determinant is multiplied by ‘p’. x1 x2 x3 Consider, D = y1 yy z1 23 z2 z3 Let D' = px1 px2 px3 y1 y2 y3

zzz 123 i.e., the elements of the 1st row of D are multiplied by ‘p’. To show that D' = p (D) Now D' = px1 (y2z3 – y3z2) – px2 (y1z3 – y3z1) + px3 (y1z2 – y2z1) = p { x1 (y2z3 – y3z2) – x2 (y1z3 – y3z1) + x3 (y1z2 – y2z1)} i.e., D' = pD Hence the theorem is proved. CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 103 Theorem 5: In each element of any row (or column) be expressed as the sum of two terms, thenthe determinant can be shown to be equal to the sum of two other determinants. x1 l1 x2 + l2 x3 + l3 Consider, y1 y2 y3 z1 z2 z3 In this determinant each element of the first row, is expressed as the sum of two terms. Expanding this determinant, we get, (x1 + l1) (y2z3 – y3z2) – (x2 + l2) (y1z3 – y3z1) + (x3 + l3) (y1 z2 – y2z1) ? { x1 (y2z3 – y3z2) – x2 (y1z3– y3z1) + x3 (y1z2 – y2z1) } ? { l1 (y2z3 – y3z2) – l2 (y1z3 – y3z1) + l3 (y1z2 – y2z1) } x xx l 1 23 l1 l2 3 = y1 y2 y3 + y1 y2 y3 z z z1 z2 3 z1 z2 3 Hence the theorem is proved. Similarly, if each element of a row or column be expressed as the sum of three terms then the determinant can be expressed as the sum of three other determinants. In this way this theorem can be extended. Theorem 6: The value of a determinant is not changed by adding to the elements of any row (or column) the same multiples of the corresponding elements of any other row (or column).

x1 x2 x3 Consider, D = yy y 12 3 z1 z2 z3 Let D' be the determinants that is obtained by adding to the elements of the first row of D, the multiples pz1, pz2, pz3 of the corresponding elements of the third row. Here ‘p’ is a certain constant. ?D= x1 + pz1 x2 + pz2 x3 + pz3 y 1 y2 y3 z 1 z2 z3 CU IDOL SELF LEARNING MATERIAL (SLM)

104 Business Mathematics and Statistics To show that D’ = D x1 x2 x3 pz1 pz2 pz3 yy Now D' = 1 y2 y3 + 1 y2 y3 By theorem 5. z1 z2 z3 z1 z2 z3 x1 x2 x3 z1 z2 z3 = y1 y2 y3 + p y1 y2 y3 By theorem 4. z1 z2 z3 z1 z2 z3 x1 x2 x3 + p × 0 (By theorem 3) = y1 yy z1 23 z2 z3 = D ?D' =D Hence the theorem is proved. By extending this theorem, we can, similarly show that, x1 + my1 + pz1 x2 + my2 pz2 x3 + my3 + pz3 yy y 12 3 z1 z2 z3 x1 x2 x3

yyy = 1 2 3 z1 z2 z3 Here m and p are constants. Example 8: Show that xy yz zx =0 yz zx xy zx xy yz CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 105 Adding the sum of the 2nd and 3rd rows to the 1st row, we get 000 = 0, since yz zx xy zx xy yz all the elements of the 1st row are zero. Example 9: Prove that z a2 bc = (a – b) (b – c) (c – a) (ab + bc + ca) b b2 ac c c2 ab Subtracting 2nd row from the 1st row and 3rd row from the second, we get a b a2 b2 bc ac b c b2 c2 ac ab c c2 ab = (a – b) (b – c) l a+b c l b+c -a c c2 ab

Subtracting 2nd row from the 1st row, we get 0 ac ac bc a (a – b) (b – c) l c2 ab c = (a – b) (b – c) (a – c) 0l l l bc a c c2 ab = (a – b) (b – c) (a – c) { – (ab + ac) + (c2 – bc – c2) } = (a – b) (b – c) (a – c) (– ab – ac – bc) = (a – b) (b – c) (c – a) (ab + bc + ca) CU IDOL SELF LEARNING MATERIAL (SLM)

106 Business Mathematics and Statistics Example 10: Prove that x y x+ z x y z+x = 4 xyz = z x+y Subtracting the sum of the second and third rows from the first row, we get 0 2z 2y yzx y = z xy 0z y y y =–2 z+x = z x+y yy y z+x = + 2z z x + y – 2y z z ? 2z { y (x + y) – yz } – 2y { yz – z (z +x) } ? 2z (xy – yz + y2) – 2y (yz – zx – z2) ? 2xyz – 2yz2 + 2y2z – 2y2z + 2xyz + 2yz2 ? 4xyz

Example 11: Prove that x + y + 2z x y y z y z + 2x = 2 (x + y + z)3 ` xz + x + 2y Let x + y + z = k, so that x + y + 2z = k + z, y + z + 2x = k + x and z + x + 2y = k + y k+z x y L.H.S. = z k+x y z x k+y CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 107 Subtracting 2nd row from the 1st row and 3rd row from the 2nd, we get kk 0 ok k n x k+y Adding 1st column to the second column k0 0 ok k = k2 l 0 0 ol l z x+z k+y z x+z k+y 5 k2 (k + y) + (x + z) 5 k2 (x + y + z + k) 5 (x + y +z)2 2 (x + y + z) 5 2 (x + y + z)2 Crammer’s Rule Cramer’s Rule is a handy way to solve for just one of the variables without having to solve the whole system of equations. The point of the Rule: instead of solving the entire system of equations, you can use Cramer’s to solve for just one single variable. Let’s use the following system of equations: 2xyz x±y±z

xyz We have the left-hand side of the system with the variables (the “coefficient matrix”) and the right-hand side with the answer values. Let DEH WKHGHWHUPLQDQWRIWKHFRHIILFLHQWPDWUL[RIWKH above system, and let DxEH WKH GHWHUPLQDQW IRUPHG E\\ UHSODFLQJ WKHx-column values with the answer-column values: CU IDOL SELF LEARNING MATERIAL (SLM)

108 Business Mathematics and Statistics System of Coefficient Answer Dx: coefficient equations matrix’s column determinant with determinant answer-columnvalues in x-column 2x 1y 1x 3 2 11 ª3º 311 1x 1y 1z 0 D1 11 «» Dx 1 11 1x 2y 1z 0 1 21 «0» 121 «» ¬0¼ 6LPLODUO\\DyDQGDzwould then be: 2 31 21 3 Dy 101 Dz 1 1 0 1 01 12 0 Evaluating each determinant (using the method explainedhere), we get: 21 1 D 1 1 1 = (–2) + (–1) + (2) – (–1) – (–4) – (1) = 3 12 1 3 11

Dx 0 1 1 =(–3)+(0)+(0)–(0)–(–6)–(0)=–3+8=3 0 21 2 31 D y 1 0 1 =(0)+(–3)+(0)–(0)–(0)–(3)–3=–6 1 01 2 13 Dz 1 1 0 =(0)+(0)+(6)–(–3)–(0)–(0)=6+3=9 1 20 Cramer’s Rule says that xD[·DyD\\·D, and zD]·D. That is: x3/3y–6/3±, and z9/3 CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 109 That’s all there is to Cramer’s Rule. To find whichever variable you want (call it “ß” or “beta”), just evaluate the determinant quotient Dß·D. Inverse of a Square Matrix Let A be a square matrix of order n. If there exists a square matrix B of order n such that AB=BA=I, then B is called the inverse of A and we denote it by A–1. If a square matrix A has inverse, then A is said to invertible. We know that A. (adj A) = (adj A). A = |A| In § ·§ · adj A adj A Ÿ ¨ ¸ =¨ ¸ A. .A=I © |A| ¹ © |A| ¹ n 1 Thus, we get A–1 =| A| . adj A A square matrix A is invertible if and only if A z 0 is non-singular, i.e. if |A| z 0. Also the inverse of a square matrix is always unique. Properties of Inverse (a) If A and B are two matrices of order n such that AB = I, then A–1 = B and B–1 = A. (b) If A and B are two non-singular matrices of order n, then (AB)–1 = B–1 . A–1 (not A–1 B–1)

(c) Inverse of inverse is the original matrix i.e., (A–1)–1 = A (d) The inverse of the transpose of a matrix is the transpose of its inverse. i.e., (A')–1 = (A–1)' (e) If A is a non-singular matrix of order n and In is the unit vector the A . A–1 = A–1. A = I. (f) The inverse of an identity matrix is the identity matrix it self. i.e., In–1 = In. GExample 1: Find the inverse of the matrix A = F I 2 –4 J H –1 3K Solution: F I 2 –4 J A=G H–1 3 K |A|= 6–4=2 0 ? A–1 exists. CU IDOL SELF LEARNING MATERIAL (SLM)

110 Business Mathematics and Statistics The cofactors of aij are C =3C12=–(–1)=1 11 C = –(–4) = 4, C22 = 2 21 I G3 1J H K= and (C ) = ij 4 2 F 3 1I H1 2K JGadj A = (Cij)’ = Inverse of A is given by 1 A–1 = . adj A | A| I1 F3 1 H4 2K 2 ? JA–1 = G FI 13 3 Example 2: Find the inverse of the matrix GG1 4 3JJ HK FI 13 4 1 33 Solution: Let A = GG1 4 3JJ H K 1 34 |A| = 1(7) – 3(1) + 3(–1) = 1 0

? A–1 exists. The cofactors are : C C13 = –1 C23=0 C 12= –1, C C33=1 11 = 7, 22 = 1, C = –3, C 21 32 = 0, C 31 = –3, 7 –1 –1 F I ? The cofactor matrix, (Cij) = G –3 1 0J GJ KH –3 0 1 7 –3 –3 F I Jadj A = (Cij)' = G –1 1 0 GJ KH –1 0 1 Inverse of A is given by A–1 = 1 . adj A | A| CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 111 F IF I 7 –3 –3 7 –3 –3 1G G JJ 1 ? A–1 = G –1 1 0 J = G –1 1 0 J H –1 0 1 K H –1 0 1 K Difference between Matrices and Determinants Matrices Determinants 1. $PDWUL[RUPDWULFHVLV D UHFWDQJXODU JULG RI $GHWHUPLQDQWLVDFRPSRQHQWRIDVTXDUHPDWUL[DQG numbers or symbols that is represented in a row it cannot be found in any other type of matrix. ... and column format. $GHWHUPLQDQWLVDQXPEHUWKDWLVDVVRFLDWHGZLWKD VTXDUHPDWUL[ 2. Matrix is the set of numbers which are covered Determinants is also set of numbers but it is covered by two brackets. by two bars. 3. It is not necessary that number of rows will be But it is necessary that number of rows will be equal equal to the number of columns in matrix. to the number of columns in determinant. 4. Matrix can be used for adding, subtracting and Determinant can be used for calculating the value multiplying the coefficients. of x, y and z with Cramer’s Rule. 5.5 Summary

We study the meaning and definition of a ‘Determinant’ along with the simple illustrative examples. We learn the method of evaluation of a determinant and solving a determinant. the six different theorems on determinants explain their usage in proving certain determinant examples. 5.6 Key Words/Abbreviations Consistence, Linear equations, Expansion, addition of determinants as explained in the theorems. 5.7 Learning Activity Study the theorems carefully and work out the examples given at the end under (MCQ and Descriptive). ................................................................................................................................................. ............................................................................................................................................... CU IDOL SELF LEARNING MATERIAL (SLM)

112 Business Mathematics and Statistics 5.8 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions = Explain the method of finding the determinant of a square matrix. = Distinguish between (a) Matrix and Determinant, (b) Minor and Cofactor = Evaluate the following: (a) 5 12 , (b) 1 –3 x2 x1 x (d) x1 , (c) 7 2 , x1 3 20 41 01 4. Evaluate the following: 1 2 34 1 32 11 1 3 1 –1 (a) 4 11 (b) 10–1 (c) 5 2 3 (d) 4 3 2 1 1 –2 5 32 1 8 31 2 1 43 3 2 14 5. Show that the following matrices are singular. F 3 4 5I F 6 I–3 2

(a) A = GG 2 –1 2JJ (b) X = GG –4 7 8 JJ H K HK –10 5 2 6 8 10 ` Are the following equations consistent? [Ans: Yes] x+y+2=0 2x + 3y + 1 = 0 x + 4 –7 = 0 ? Find k so that the following equations may be simultaneously true x + ky – 3 = 0 (2k + x) + y – 5 = 10 3x – y + 11 = 0 [Ans: k = 3] 8. Prove that xyz 2x 2x = (x + y + z)3 2y y z z 2y 2z 2z z x y CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 113 9. Show that 1 11 x yz = (x – y) (y – z) (z – x) x2 y2 z2 10. Prove that  y + z x z x y  z+x  = 8xyz yz yx  z y z x x+y  11. Show that 1 w w2 = 0 and also w w2 1 w2 1 w 1 w3 w2 =3 w3 1 w w2 w 1 where w is one of the imaginary cube roots of unity. 12. Prove that x+1 x+m x+n

y+1 y+m y+n =0 z+1 z+m z+n 13. Show that x2 + 1 2x + 1 3 = (x – 1)3 2x + 1 x + 2 3 1 11 14. Show that (y + z)2 x2 x2 y2 (z + x)2 y 2 = 2xyz (x + y + z)3 (x + y)2 z2 z2 CU IDOL SELF LEARNING MATERIAL (SLM)

114 Business Mathematics and Statistics 15. If n+1 m n m l+m n l n a a2 a3 + 1 b b2 b3 + 1 = 0, show that abc = – 1 `A c c2 c3 + 1 r e 16. Solve t 1 +x 1 x 1 x h e 1x1x1x =0 1 x 1 x 1x f o 17. Solve l l 1x 3 o w 2x 4 3x 2 =2 i n 2x 6 x g = Prove that e q x y z z+y u x + z y z x = (x + y + z) (x2 + y2 + z2) a t x y y+x z i o 19. Show that lm l n s m+n

consistent? y – x + z = 0 3x + 7y – 3z = 0 4x + y + z = 0 [Ans: x = 0 , x = 3] [Ans: x = 2, – or 3]

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Determinants 115 = Without directly expanding the determinant, but using only the well-known properties, prove that ab l a a 1b bc l b = b 1c ca l c c 1a 22. Solve x+a b c =0 c x+b a ab x+c 23. Prove that 2 2 a 2ab b 2ab b2 a2 is a perfect square. 24. Without actual expansion, but by using properties of determinants, prove that a2 a bc a3 a2 l

2 b ca = 3 2 l b b b c2 c ab c3 c2 l 25. Solve for x lll a x a =0 bb x 26. Show that a2 a l b2 b l = – (a – b) (b – c) (c – a) c2 c l CU IDOL SELF LEARNING MATERIAL (SLM)

116 Business Mathematics and Statistics 27. Express the square of the determinant a1 b1 0 =2 b2 0 = 3 b3 0 as a third order determinant. Now obtain the value of this third order determinant. 28. Find the value of the determinant ab0 c0a 0c b and then express its square as a third order determinant. What is its value? 29. Write down the determinant. ap + bq ar + bs cp + dq cr + ds as the sum of four determinants and hence prove that it equals to the product of the two determinants, namely

ab pq × cd rs ` Solve the following system of equation by Cramer’s Rule. 2x + y + z = 7 3x – y – z = –2 x + 2y – 3z = – 4 5.9 References References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

Permutations and Combinations 117 UNIT 6 PERMUTATIONS AND COMBINATIONS Structure 6.0 Learning Objectives 6.1 Introduction 6.2 Combinations 6.3 Interpretation of O! 6.4 Distribution in Groups 6.5 Illustrative Examples 6.6 Summary 6.7 Key Words/Abbreviations

6.8 Learning Activity 6.9 Unit End Questions (MCQ and Descriptive) 6.10 References 6.0 Learning Objectives After studying this unit, you will be able to:  Explain the meanings of ‘permutation’ and ‘combination’.   Elaborate the notation nPr and nCr and the related theorem for finding the value of nPr and nCr where r d n.  Describe the interpretation of nCo, nCn, O! and the rules and corollories.   Illustrate the method of solving related examples with reference to illustration examples.

CU IDOL SELF LEARNING MATERIAL (SLM) 118 Business Mathematics and Statistics 6.1 Introduction In the modern world of business, commerce and industry, we come across certain practical problems which involve a number of different arrangements (Permutations) and selections (Combinations) of men, materials and things. Major business decisions are more often dependent upon a wise and judicious choice of the different arrangements or selections. The success of such decisions depends upon the thorough study of the different permutations and combinations. A study of the theoretical principles and problems of permutations and combinations is not only amusing but also useful. For instance, if there are two vacancies of salesmen in a firm, i.e., a vacancy of a junior salesman and that of a senior salesman and the management interviews three applicants A, B, C then the management can take anyone of the following six possible decisions in filling up the two vacancies. These six possible arrangements, that is, AB, BA, BC, CB, CA and AC are known as the ‘permutations’ of three applicants A, B, C taken two at a time. Junior Salesman Senior Salesman AB BA

BC CB CA AC Thus the different arrangements that ·can be made by taking some or all of a number of things, are known as ‘Permutations’. To give another example, if three different numbers, i.e., 2, 5 and 6 are given and we are required to find the number of permutations of these three numbers taking two at a time then the following would be the different permutations. 26, 56, 62, 65, 26 On the other hand, if we are required to find the number of permutations of the three numbers 2, 5, 6 taking all the three numbers at a time, then the permutations would be: 256, 562, 625, 526, 652, 265

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Permutations and Combinations 119 Narration: P is a notation that is used to denote the number of permutations of n different things, when r things are taken. at a time. For example, if 2 things are taken at a time out of 3 different things, then the number of different permutations would be 3P2 and in this case 3P2 = 6. There is a fundamental theorem which forms the basis of certain general propositions in Permutations. This theorem can be stated as: If there are m ways of performing an operation and after the operation has been performed in anyone of the m ways, if a second operation can be performed in n ways then the number of ways of performing the two operations would be mn. To show that the number of permutations of n different things taken r at a time is n (n – 1) (n – 2) .................. (n – r + 1) Suppose there are r different blank spaces, each of which has to be filled up by any of the n different things. 12 3 ........ r (n – r + 1) (n) (n – 1) (n – 2) There are n ways to filling up the first space and when it has been filled up by anyone of the n things, then there being (n – 1) things remaining, it follows that the second space can be filled up in (n – 1) ways. Similarly the third space can be filled up in (n – 2) ways and proceeding in, that manner it is clear that the rth space can be filled up in (n – r + 1) ways. Therefore, the number of ways of filling up all the r blank spaces, would be: