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

$$\begin{gathered} \frac{\left(2n\right)!}{3!(2n-3)!}:\frac{n!}{2!(n-2)!}=44:3 \\ \Rightarrow \frac{\left(2n\right)!}{3!(2n-3)!}\times \frac{2!(n-2)!}{n!}=\frac{44}{3} \\ \Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)\left[\right(2n-3)!]}{3(2!)(2n-3)!}\times \frac{2!(n-2)!}{n(n-1)\left[\right(n-2)!]}=\frac{44}{3} \\ \Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)}{3}\times \frac{1}{n(n-1)}=\frac{44}{3} \\ \Rightarrow \frac{\left(2n\right)(2n-1)\left(2\right)(n-1)}{3}\times \frac{1}{n(n-1)}=\frac{44}{3} \\ \Rightarrow \frac{4(2n-1)n(n-1)}{3}\times \frac{1}{n(n-1)}=\frac{44}{3} \\ \Rightarrow \frac{4(2n-1)}{3}=\frac{44}{3} \\ \Rightarrow (2n-1)=11 \\ \Rightarrow 2n=12 \\ \Rightarrow n=6 \end{gathered}$$