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: $\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}$
Observe that in each term of above summation, sum of the arguements of the factorial in the denominator is constant for each term and equal to $n-2$, so you should try to write each term as a binomial coefficient.
And since the constant is $n-2$ we are motivated to write the summation as,
$n(n-1)\sum_{k=2}^n \frac{(n-2)!}{(n-k)!(k-2)!}=n(n-1)\sum_{r=0}^{n-2} \dbinom{n-2}{r}$
Now as mentioned in the other posts $\sum_{r=0}^{n-2} \dbinom{n-2}{r}=2^{n-2}$
So, you are done.
A: Note that $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!} = \sum_{k=0}^n k(k-1) {n \choose k}.$$
Thus, consider $f$ defined by $f(x) = (1+x)^{n}$. Then
$$f(x) = \sum_{k=0}^n {n \choose k} x^k.$$
Then $$f''(x) = \sum_{k=2}^n k(k-1) {n \choose k} x^{k-2},$$
and
$$f''(x) = n(n-1)(1+x)^{n-2}.$$
All you have to do is take $x=1$.
A: 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 $x=1$.
A: 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, $\sum_{k=2}^n\frac{n!}{(n-k)!(k-2)!}=\sum_{k=2}^n2\binom{n}{k}\binom{k}{2}=\binom{n}{2}2^{n-1}$
