Permutation and Combination – Exercise 4.6 – Class X

Previous exercise – Permutation and Combination – Exercise 4.5

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Permutation and Combination – Exercise 4.6

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

(i) if ⁿC₄ = ⁿC₇ find n

(ii) IF ⁿP_r = 840, ⁿC_r = 35 find n

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

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Permutation and Combination – Exercise 4.6 – Solutions:

1. Evaluate

(i) ¹⁰C₃

(ii)⁶⁰C₆₀

(iii)¹⁰⁰C₉₇

Solution:

(i) ¹⁰C₃

We know, ⁿC_r = ^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, ⁿC_r = ^n!/_{(n – r)!r!}

⁶⁰C₆₀ = ^60!/_{(60 – 60)!60!}

= ^60!/_{0! x 60!}

= ^60!/_{1 x 60!}

= 1

 

(iii)¹⁰⁰C₉₇

We know, ⁿC_r = ^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

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2. (i) if ⁿC₄ = ⁿC₇ find n

(ii) IF ⁿP_r = 840, ⁿC_r = 35 find n

Solution:

(i) if ⁿC₄ = ⁿC₇ , we have to find n

ⁿC₄ = ⁿC₇

ⁿC₄ = ⁿC_{n – 7}

4 = n – 7

n = 4 + 7 = 11

 

(ii) If ⁿP_r = 840, ⁿC_r = 35, we have to find n

We know, ⁿP_r =r! ⁿC_r

840 = r! x 35

r! = ⁸⁴⁰/₃₅ = 24

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

Therefore, r = 4

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3. 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

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4. 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.

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Next exercise Permutation and Combination – Exercise 4.7

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