Question is to Evaluate $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!} $$
What i have done so far is $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}=n(n-1)\sum_{k=2}^n \frac{(n-2)!}{(n-k)!(k-2)!}=n(n-1)\sum_{k=2}^n \binom{n-2}{k-2}$$
I could see that $$\sum_{k=2}^n \binom{n-2}{k-2}$$looks much similar to $$\sum_{k=0}^n \binom{n}{k}=(1+1)^n$$
So, we should have something like $$\sum_{k=2}^n \binom{n-2}{k-2}=(1+1)^{n-2}$$
But i am not very sure if this is even true.
I tried adding $\binom{n-2}{k-2}$ with $\binom{n-2}{k-1}$ with a formula $\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{k}$ hoping to reduce $n-2$ in the expansion to $n$ but is was not helpful.
I tried to consider $(1+x)^n$ expansion and tried differentiating it but this was also not very helpful.
I would be thankful if some one can help me to clear this.
Thank you