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

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

$\begin{array}{l}\left.\frac{{ }^{n+2}{C}_8}{{ }^{n-2}{P}_4}=\frac{57}{16}\left[{\because }^n{C}_r=\frac{n!}{r!(n-r)!}\right]{\text{ and }}^n{P}_r=\frac{n!}{(n-r)!}\right]\\ \Rightarrow \frac{(n+2)!}{8!(n+2-8)!}\times \frac{(n-2-4)!}{(n-2)!}=\frac{57}{16}\\ \Rightarrow \frac{(n+2)(n+1)n\cdot (n-1)}{\mathrm{8.7.6.5}\cdot \mathrm{4.3.2.1}}=\frac{57}{16}\\ \Rightarrow (n+2)(n+1)n(n-1)=143640\\ \Rightarrow \left(n^2+n-2\right)\left(n^2+n\right)=143640\\ \Rightarrow {\left(n^2+n\right)}^2-2\left(n^2+n\right)+1=143640+1\\ \begin{array}{l}\Rightarrow {\left({\mathrm{n}}^2+\mathrm{n}-1\right)}^2=(379{)}^2\\ \Rightarrow {\mathrm{n}}^2+\mathrm{n}-1=379\left[\because {\mathrm{n}}^2+\mathrm{n}-1>0\right]\\ \Rightarrow {\mathrm{n}}^2+\mathrm{n}-1-379=0\\ \Rightarrow {\mathrm{n}}^2+\mathrm{n}-380=0\\ \Rightarrow (\mathrm{n}+20)(\mathrm{n}-19)=0\\ \Rightarrow \mathrm{n}=-20,\mathrm{n}=19\end{array}\end{array}$

$\because$ n is not negative.

$\therefore$ n = 19