Official solution
\[{{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11\]
\[ \Rightarrow \] \[{{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}\]
\[ \Rightarrow \] (n + 2)! = 11.6! (n - 2)!
\[ \Rightarrow \] (n + 2) (n + 1) n (n - 1) = 11.6!
\[ \Rightarrow \] (n + 2) (n + 1) n (n - 1) = 11 . 6 . 5 . 4 . 3 . 2 . 1
\[ \Rightarrow \] (n + 2) (n + 1) n (n - 1) = 11 . 10 . 9 . 8
\[ \therefore \] n = 9
This value of n satisfy the equation,
n2 + 3n - 108 = 0