If ${}^{n+2}{C}_8:{}^{n-2}{P}_4=\frac{57}{16}$ then $n$ is equal to ?

The correct option is A

$19$

Explanation for correct option

Given, ${}^{n+2}{C}_8:{}^{n-2}{P}_4=\frac{57}{16}$

Simplifying the above we get

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{\left(n+2\right)!}{8!\left(n+2-8\right)!}}}{{\displaystyle \frac{\left(n-2\right)!}{\left(n-2-4\right)!}}}=\frac{57}{16}\left[\because {}^{n}C_m=\frac{n!}{m!\left(n-m\right)!},{}^{n}P_m=\frac{n!}{\left(n-m\right)!}\right] \\ \Rightarrow \frac{{\displaystyle \frac{\left(n+2\right)!}{8!\left(n-6\right)!}}}{{\displaystyle \frac{\left(n-2\right)!}{\left(n-6\right)!}}}=\frac{57}{16} \\ \Rightarrow \frac{\left(n-1\right)\left(n\right)\left(n+1\right)\left(n+2\right)}{8!}=\frac{19\times 3}{8\times 2} \\ \Rightarrow \left(n-1\right)\left(n\right)\left(n+1\right)\left(n+2\right)=\frac{19\times 3\times 7!}{2} \\ \Rightarrow \left(n-1\right)\left(n\right)\left(n+1\right)\left(n+2\right)=\frac{1\times 2\times 3\times 4\times 5\times 6\times 7\times 3\times 19}{2} \\ \Rightarrow \left(n-1\right)\left(n\right)\left(n+1\right)\left(n+2\right)=21\times 20\times 19\times 18 \end{gathered}$$

Therefore clearly $n=19$

Hence, option(A) i.e. $19$ is correct