## Algebra 2 (1st Edition)

$n=10$
$_nP_r=\frac{n!}{(n-r)!}$, hence here the equation is: $\frac{n!}{(n-6)!}=5\frac{n!}{(n-5)!}$ If we multiply both sides by $(n-5)!=(n-5)\cdot(n-6)!$ we get: $n!(n-5)=5n!\\n-5=5\\n=10$