Find the sum $\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}.$ Find the following sum $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}.$$I found that ,  $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}=n!\left[\frac{1}{(n-2)!0!}+\frac{1}{(n-3)!1!}+\cdots +\frac{1}{0!(n-2)!}\right]$$From here how I proceed ?

Outline: By the Binomial Theorem we have
$$(1+x)^n=\sum_{k=0}^n \frac{n!}{(n-k)!k!}x^k.$$
Differentiate twice and set $x=1$.
