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 ?

A combinatorial aproach:
$\dfrac{n!}{(n-k)!(k-2)!}=(k-1)(k)\displaystyle\binom{n}{k}=2\binom{n}{k}\binom{k}{2}$. But the last can be thinking like total ways to select a team with $k$ elements from a set of $n$ and then select two captains from this $k$, which is the same that select first the two captains and then the other $k-2$: $\binom{n}{k}\binom{k}{2}=\binom{n}{2}\binom{n-2}{k-2}$. Then: 
$$\displaystyle\sum_{k=2}^n\binom{n}{k}\binom{k}{2}=\displaystyle\sum_{k=2}^n\binom{n}{2}\binom{n-2}{k-2}=\binom{n}{2}\displaystyle\sum_{k=0}^{n-2}\binom{n-2}{k}=\binom{n}{2}2^{n-2}$$
Finally, $\sum_{k=2}^n\frac{n!}{(n-k)!(k-2)!}=\sum_{k=2}^n2\binom{n}{k}\binom{k}{2}=\binom{n}{2}2^{n-1}$
