Thanks for those who helped me yesterday. I have another one and I'd appreciate if you can shed some light upon to keep me going. This one is related with binomial theorem. $$\sum_{k=1}^m k(k-1){m\choose k} = m(m-1) 2^{m-2}$$
Thanks for your thoughts!
Both sides are all ways of choosing a subset of $m$ things, with $2$ distinguished things in the subset (forcing the size of the subset to be at least $2$). The left side is conditioning on the size of the subset.
Think of it as choosing a committee from $m$ people, where there is a President and Vice-President who are distinct people. On the left hand side, you choose a committee size of $k$ people, then from those $k$ people choose the President and Vice-President. By rule of product this gives you $k(k-1)\binom{m}{k}$. By rule of sum we can also see that the size of the committee partitions the set of all committees, so the total way to choose such a committee is $\sum_{k = 1}^mk(k-1)\binom{m}{k}$.
On the right hand side, you first choose the President and Vice-President from $m$ people, and from the remaining $m-2$ people, choose whether or not they are on the committee, which is $2^{m-2}$ ways. By rule of product the total number of ways to choose a committee is $m(m-1)2^{m-2}$. By counting in two ways, the two sides are equal.
$$\sum_{k=1}^m {m\choose k}x^k = (1+x)^m$$ differentiating with respect to $x$ $$\sum_{k=1}^m k{m\choose k}x^{k-1}=m(1+x)^{m-1}$$
differentiate once more with respect to $x$ $$\sum_{k=1}^m k(k-1){m\choose k}x^{k-2}=m(m-1)(1+x)^{m-2}$$ when $x=1$ $$\sum_{k=1}^m k(k-1){m\choose k}=m(m-1)2^{m-2}$$
Notice that your sum is $2 \sum_k {k\choose 2} {m\choose k}$ (I am deliberately omitting the range for $k$; why?).
This suggests that we might consider the more general sum:
$$S_{m,l} := \sum_k {k\choose l} {m\choose k}$$
Here are my three favorite ways to compute this at a glance:
1) $S_{m,l}$ is the number of ways in which we can divide a group of $m$ people into three teams—A, B, and C—such that team C has exactly $l$ members.
2) $S_{m,l}$ is the coefficient of $x^l$ in $(1+(1+x))^m = (2+x)^m$.
3) $S_{m,l}$ is $\frac{1}{l!}$ times the $l^{\operatorname{th}}$ derivative of $(1+x)^m$, evaluated at $x=1$.
In any case, we observe: $$S_{m,l} = {m\choose l} 2^{m-l}$$
In particular, your sum is $2 S_{m,2} = 2 {m\choose 2} 2^{m-2} = m(m-1)2^{m-2}$.
While there might be little need to adding to many good answers, here is yet another point of view. One has for all $k,m\in \Bbb N$ $$ \binom k2\binom mk=\binom m{2,k-2,m-k}=\binom m2\binom{m-2}{k-2} $$ which counts the ways to partition $m$ objects into labelled groups of $2$, $k-2$, and $m-k$ elements. Now multiply by $2$ and sum over all $k$ to get your formula.
Consider the polynomial $P(x) = (x + 1)^m$. Use the binomial theorem, derive twice and evaluate at $x=1$. You get LHS. Now, derive twice and evaluate at $x= 1$ the factorized form of $P(x)$. You get RHS.
