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

Question

Let $n ≥ 2$. Evaluate:

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

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unable to solve this series problem.

Answer 1

Differentiating the both sides of $$(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$$ will give you $$n(1+x)^{n-1}=\sum_{k=1}^{n}\binom{n}{k}kx^{k-1}.$$ Differentiating the both sides of this equation will give you $$n(n-1)(1+x)^{n-2}=\sum_{k=2}^{n}\binom{n}{k}k(k-1)x^{k-2}.$$

Hence, setting $x=1$ gives you $$\begin{align}\sum_{k=2}^{n}\frac{n!}{(n-k)!(k-2)!}&=\sum_{k=2}^{n}\frac{n!}{(n-k)!k!}\cdot k(k-1)\\&=\sum_{k=2}^{n}\binom{n}{k}k(k-1)\\&=n(n-1)(1+1)^{n-2}\\&=n(n-1)\cdot 2^{n-2}.\end{align}$$

Answer 2

Hint1:

$$\sum_{k=2}^{n}\frac{n!}{\left(n-k\right)!\left(k-2\right)!}=n\left(n-1\right)\sum_{k=0}^{n-2}\binom{n-2}{k}$$

Hint2:

$$\left(p+q\right)^{m}=\sum_{k=0}^{m}\binom{m}{k}p^{k}q^{m-k}$$ What happens if $p=q=1$?