If $p={}^{n+2}{P}_{n+2};q={}^n{P}_{11};r={}^{n-11}{P}_{n-11}$ and if $P=182qr$, then $\text{'}n\text{'}=$

Given $p={}^{n+2}{P}_{n+2}=(n+2)!$

$q={}^n{P}_{11}=\frac{n!}{(n-11)!}$

and $r={}^{n-11}{P}_{n-11}=(n-11)!$

Given relation is $p=182qr$

$(n+2)!=182\left(\frac{n!}{(n-11)!}\right)(n-11)!$

$(n+2)!=182(n!)$

$\frac{(n+2)!}{n!}=182\Rightarrow \frac{(n+2)(n+2)n!}{n!}=182$

$\Rightarrow n^2+3n+2=182\Rightarrow n^{2}3n-180=0\Rightarrow n^2+15n-12n-180=0$

$\Rightarrow (n+15)(n-12)=0$

$\Rightarrow n=12,n=-15$ (not possible)

$\therefore n=12$