Evaluate: $\sum_{k=2}^n\frac {n!}{(n-k)!(k-2)!}$

Question

Evaluate:

$$\sum_{k=2}^n\frac {n!}{(n-k)!(k-2)!}$$

Attempt

$S_2=\frac {n!}{(n-2)!}$

$S_3=\frac {n!}{(n-3)!}$

$S_4=\frac {n!}{2(n-4)!}$

$\vdots$

$S_{n-1}=\frac {n!}{1!(n-3)!}$

$S_n=\frac {n!}{(n-2)!}=\frac{(n-2)!(n-1)n}{(n-2)!}=n(n-1)$

Is it correct? If wrong, Kindly quote where is the mistake?

Answer 1

$$\frac{n!}{(n-k)!(k-2)!}=n(n-1)\frac{(n-2)!}{\{n-2-(k-2)\}!\cdot(k-2)!}=n(n-1)\binom{n-2}{k-2}$$

$$\sum_{k=2}^n\frac{n!}{(n-k)!(k-2)!}=n(n-1)\sum_{k=2}^n\binom{n-2}{k-2}$$

Now set $k-2=r$ to get $$\sum_{k=2}^n\binom{n-2}{k-2}=\sum_{r=0}^{n-2}\binom{n-2}r=(1+1)^{n-2}$$

Answer 2

This is $\sum k(k-1)\binom{n}{k}$ which is $f''(1)$ where $f(x)=\sum\binom{n}{k}x^k=(1+x)^n$

Answer 3

after your method i got $$2^{n-2}(n-1)n$$