Previous exercise – Permutation and Combination – Exercise 4.5
Permutation and Combination – Exercise 4.6
- Evaluate (i) 10C3 (ii)60C60 (iii)100C97
(i) if nC4 = nC7 find n
(ii) IF nPr = 840, nCr = 35 find n
- If 2nC3 : nC3 = 11 : 1 , find n
- Verify that 8C4 + 8C5 = 9C4
- Prove that nCr/n-1Cr-1 = n/r when 1 ≤ r ≤ n
Permutation and Combination – Exercise 4.6 – Solutions:
- 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
- If 2nC3 : nC3 = 11 : 1 , find n
Solution:
2nC3 : nC3 = 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
- 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