TITLE: Combinatorial proof of $n(n-1)2^{n-2} = \sum_{i=2}^{n}k(k-1)\binom{n}{k}$
QUESTION [5 upvotes]: Today I had a combinatronics test, and I really couldn't do this problem, it said to give a combinatorial proof of:
$$ n(n-1)2^{n-2} = \sum_{i=2}^{n}k(k-1)\binom{n}{k} $$
It said that it should be derived from the binomial theorem. But I am not able to see it. I thought about using the identity 
$$2^{n} = \sum_{i=0}^{n}\binom{n}{k}$$
Also I thought about using the fact that:
$$1 +2+ 3+ \cdots + n = \frac{n(n+1)}{2}$$
But I didn't get anywhere.
It MUST be a combinatorial proof.
Thanks in advance for your help.

REPLY [2 votes]: Let $$f(x)=(x+1)^n$$
$$f(x)=(x+1)^n=\sum_{k=0}^{n}x^k\binom{n}{k}$$
$$f''(x)=n(n-1)(x+1)^{n-2}=\sum_{k=2}^{n}k(k-1)x^{k-2}\binom{n}{k}$$
$$f''(1)=n(n-1)2^{n-2}=\sum_{k=2}^{n}k(k-1)\binom{n}{k}$$