Question

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

Answer 1

Given,

$\Rightarrow \frac{\mathrm{4.3.2}!(n-4)!}{2!(n-2)(n-3)(n-4)!}=\frac{2}{1}$

$\Rightarrow \frac{4-3}{(n-2)(n-3)}=\frac{2}{1}$

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

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

But for $0$ $(n-2)!$ and $(n-4)!$ are not meaningfully so $n=5.$

Hence, the answer is $5.$