Question

Find $n$, if:
$\frac{n!}{3!(n-3)!}:\frac{n!}{5!(n-5)!}=5:3$

Answer 1

$\frac{n!}{3!n-3!}.\frac{5!n-5!}{n!}=\frac{5}{3}$

$\frac{4\times 5}{(n-4)(n-3)}=\frac{5}{3}$

$3\times 4=(n-4)(n-3)$

$\Rightarrow n-4=3$

$\Rightarrow n=7$