Question

If ${}^{2n}{C}_3:{}^n{C}_2$ = 44:3, then for which of the following values of r, the value of ${}^n{C}_r$ will be 15

• A. r=3
• B. r=4
• C. r=6
• D. r=5

Answer 1

The correct option is B

r=4

$\frac{\left(2n\right)!}{(2n-3)!.3!}\times \frac{2!\times (n-1)!}{n!}$ = $\frac{44}{3}$

$\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{3n(n-1)}$ = $\frac{44}{3}$

$\Rightarrow 4(2n-1)$ = $\Rightarrow 2n$ = $12\Rightarrow n$ = 6

Now ${}^6{C}_r$ = $15\Rightarrow {}^6{C}_r$ = ${}^6{C}_2$ or ${}^6{C}_4\Rightarrow r$ = 2,4