Question

If $\frac{\left(2n\right)!}{3!(2n-3)!}and\frac{n!}{2!(n-2)!}$ are in the ratio 44:3, find n.

Answer 1

We have,
$\frac{\frac{\left(2n\right)!}{3!(2n-3)!}}{\frac{n!}{2!(n-2)!}}=\frac{44}{3}\Rightarrow \frac{\left(2n\right)!\times 2!(n-2)!}{3!(2n-3)!\times n!}=\frac{44}{3}\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)(2n-3)!\times 2!(n-2)!}{3\times 2!(2n-3)!\times n(n-1)(n-2)!}=\frac{44}{3}\Rightarrow \frac{2n(2n-1)(2n-2)}{3n(n-1)}=\frac{44}{3}\Rightarrow \frac{2(2n-1)\times 2(n-1)}{3(n-1)}=\frac{44}{3}\Rightarrow 4(2n-1)=44\Rightarrow 2n-1=11\Rightarrow 2n=12\Rightarrow n=6\therefore n=6$