Question

If ${2n_C}_4:{n_C}_3=21:1$, then find the value of n.

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The correct option is B 5

$\frac{{2n_C}_4}{{n_C}_3}=\frac{21}{1}$

${2n_C}_4=21{n_C}_3$

$\frac{\left(2n\right)!}{(2n-4)!4!}=21\frac{n!}{(n-3)!3!}$

$2n(2n-1)(2n-2)(2n-3)=84n(n-1)(n-2)$

$4(2n-1)(2n-3)=84(n-2)$

From the given options, $n=5$ satisfies the above equation.

Hence, option B is correct.