$\sum_{k=1}^m k(k-1){m\choose k} = m(m-1) 2^{m-2}$ 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!

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. 
