Question

If $\frac{\left(2n\right)!}{(3!)\times (2n-3)!}:\frac{n!}{(2!)\times (n-2)!}=44:3,$ find the value of n.

Answer 1

$\frac{\left(2n\right)(2n-1)(2n-2)\times (2n-3)!}{6\times (2n-3)!}\times \frac{2\times (n-2)!}{n(n-1)\times (n-2)!}=\frac{44}{3}$

$\Rightarrow \frac{8\times n\times (n-1)\times (2n-1)}{6\times n\times (n-1)}=\frac{44}{3}\Rightarrow 2n-1=\frac{44}{3}\times \frac{6}{8}\Rightarrow 2n-1=11\Rightarrow n=6.$