Question

Determine n if ^2nC3: ^nC3 = 9:1.

(a) 7

(b) 14

(c) 28

(d) 32

I have been asked this question in a job interview.

I'm obligated to ask this question of Combinations topic in portion Permutations and Combinations of Mathematics – Class 11

Answer 1

Correct answer is (b) 14

The best I can explain: We know, ^nCr = \(\frac{n!}{(n-r)! r!}\).

^2nC3 / ^nC3 = \(\frac{\frac{n!}{3! (2n-3)!}}{\frac{n!}{3! (n-3)!}}\)

= \(\frac{2n! (n-3)!}{n! (2n-3)!}\)

9/1 = \(\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}\)

9 = \(\frac{2(2n-1)2}{(n-2)}\)

9n-18 = 8n-4

=>n=14.