Permutation and Combination – Exercise 4.6 – Class X

Previous exercise – Permutation and Combination – Exercise 4.5


Permutation and Combination – Exercise 4.6

  1. Evaluate (i) 10C3 (ii)60C60 (iii)100C97

(i) if nC4 = nC7 find n

(ii) IF nPr = 840, nCr = 35 find n

  1. If 2nC3 : nC3 = 11 : 1 , find n
  2. Verify that 8C4 + 8C5 = 9C4
  3. Prove that nCr/n-1Cr-1 = n/r when 1 ≤ r ≤ n

Permutation and Combination – Exercise 4.6 – Solutions:

  1. Evaluate

(i) 10C3

(ii)60C60

(iii)100C97

Solution:

(i) 10C3

We know, nCr = n!/(n – r)!r!

10C3 = 10!/(10 – 3)!3!

= 10!/7! x 3!

= 7! x 8 x 9 x 10/7! x 2 x 3

= 8 x 9 x 10/2 x 3

= 4 x 3 x 10

= 120

 

(ii)60C60

We know, nCr = n!/(n – r)!r!

60C60 = 60!/(60 – 60)!60!

= 60!/0! x 60!

= 60!/1 x 60!

= 1

 

(iii)100C97

We know, nCr = n!/(n – r)!r!

100C97 = 100!/(100 – 97)!97!

= 10!/3! x 97!

= 97! x 98 x 99 x 100/97! x 2 x 3

= 98 x 99 x 100/2 x 3

= 98 x 33 x 50

= 161700


2. (i) if nC4 = nC7 find n

(ii) IF nPr = 840, nCr = 35 find n

Solution:

(i) if nC4 = nC7 , we have to find n

nC4 = nC7

nC4 = nCn – 7

4 = n – 7

n = 4 + 7 = 11

 

(ii) If nPr = 840, nCr = 35, we have to find n

We know, nPr =r! nCr

840 = r! x 35

r! = 840/35 = 24

r! = 1 x 2 x 3 x 4 = 4!

Therefore, r = 4


  1. If 2nC3 : nC3 = 11 : 1 , find n

Solution:

2nC3nC3 = 11 : 1

2nC3/nC3 = 11/1

2nC3 = 11 x nC3

(2n)!/(2n-3)! 3! = 11 x n!/(n – 3)!3!

(2n-3)! (2n – 2)(2n – 1) 2n/(2n – 3)! 3! = 11 x (n-3)!(n-2)(n-1)n/(n-3)! 3!

(2n – 2)(2n – 1) 2n/ 3! = 11 x (n-2)(n-1)n/3!

(2n – 2)(2n – 1) 2n = 11 x (n-2)(n-1)n

2(n-1) x 2 x n x(2n-1) = 11n(n-2)(n-1)

4(2n-1) = 11(n – 2)

8n – 4 = 11n – 22

8n – 11n = 4 – 22

-3n = – 18

n = 18/3 = 6


  1. Verify that 8C4 + 8C5 = 9C4

Solution:

LHS = 8C4 + 8C5

= 8!/(8-4)!4! + 8!/(8-5)!5!

= 8!/4!4! +8!/3!5!

= 4! x 5 x 6 x 7 x 8/4! x 1 x 2 x 3 x 4 + 5! x 6 x 7 x 8/5! x 1 x 2 x 3

= 5 x 7 x 2 + 7 x 8

= 70 + 56

RHS = 9C4

= 9!/(9-4)!4!

= 9!/5!4!

= 5! x 6 x 7 x 8 x 9/5! x 1 x 2 x 3 x 4

= 7 x 2 x 9

= 126

LHS = RHS

Hence proved.


Next exercise Permutation and Combination – Exercise 4.7


 

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