Question

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$, find the value of $n$?

Answer 1

$\frac{\frac{n!}{2!(n-2)!}}{\frac{n!}{4!(n-4)!}}=\frac{2}{1}$

$\Rightarrow \frac{4!(n-4)!}{2!(n-2)!}=\frac{2}{1}$

$\Rightarrow \frac{12}{(n-2)(n-3)}=2$

$\Rightarrow (n-2)(n-3)=6$

$\Rightarrow n^2-5n+6=6$

$\Rightarrow n^2-5n=0$

$\Rightarrow n(n-5)=0$

$\Rightarrow n=0$ or $n=5$

$(n-2)!$ won't be defined.