Solution:

$ \begin{aligned} \frac{n !}{3 !(n-3) !}+\frac{(n+2) !}{3 !(n-1) !} =\frac{n !}{(n-3) !} \\\\ \frac{n(n-1)(n-2)}{6}+\frac{(n+2)(n+1)(n)}{6} =n(n-1)(n-2) \\\\ (n-1)(n-2)+(n+2)(n+1) =6(n-1)(n-2) \\\\ (n+2)(n+1) =5(n-1)(n-2) \\\\ n^{2}+3 n+2 =5\left(n^{2}-3 n+2\right) \\\\ 4 n^{2}-18 n+8 =0 \\\\ 2 n^{2}-9 n+4 =0 \\\\ (2 n-1)(n-4) =0 \\\\ n =\frac{1}{2}, 4 \\\\ But, n \neq \frac{1}{2} \\\\ Hence, n =4 \\ \end{aligned} $