Question

If ${}^{n+2}{C}_8{:}^{n-2}{P}_4=57:16$ , then the value of $n$ is

Answer 1

Answer: 19

$\frac{\left(n+2\right)!}{8!\left(n-6\right)!}\times \frac{\left(n-6\right)!}{\left(n-2\right)!}=\frac{57}{16}$
$\Rightarrow \left(n+2\right)\left(n+1\right)n\left(n-1\right)=\frac{57}{16}\times 8!$
$\Rightarrow \left(n-1\right)n\left(n+1\right)\left(n+2\right)=57\times 7\times 6\times 5\times 4\times 3$
$\Rightarrow \left(n-1\right)n\left(n+1\right)\left(n+2\right)=18\times 19\times 20\times 21$
$\Rightarrow n=19$