If ${ }^{n+5}{p}_{n+1}=\frac{1}{2}\left(11\right)(n-1)\left({ }^{n+3}{p}_n\right)$ then $n$ is equal to

$\frac{1}{2}\left(11\right)(n-1)=\frac{{ }^{n+5}{P}_{n+1}}{{ }^{n+3}{P}_n}$

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

$\Rightarrow \frac{1}{2}\left(11\right)(n-1)=\frac{1}{4}(n+5)(n+4)$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}22n-22=n^2+9n+20\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{n}^2-13n+42=0\Rightarrow n=6,7\end{array}$