Problem: Find a simple expression for $ 0\binom{m}{0}\binom{n}{k}+1\binom{m}{1}\binom{n}{k-1}+\cdots+k\binom{m}{k}\binom{n}{0} . $ Solution: This is just $\sum_{i = 0}^ki\binom{m}{i}\binom{n}{k-i}$ which is close to Vandermonde's identity which is just $\sum_{i = 0}^k\binom{m}{i}\binom{n}{k-i} = \binom{m+n}{k}$. In fact, we can just slightly adjust the double counting argument to kill this problem. Suppose we have $m$ cats and $n$ dogs and we want to create a group of $k$ animals from among the $m+n$ animals and appoint one of the kitties to be the leader. There are $\sum_{i = 0}^ki\binom{m}{i}\binom{n}{k-i}$ such ways to do this by casing on how many cats there will be in the group. On the other hand, if we just select a kitty to be the leader ($m$ options), we have $m+n-1$ animals leftover and we can just choose $k-1$ from among them to add to the group, so we conclude $ \sum_{i = 0}^ki\binom{m}{i}\binom{n}{k-i} = m\binom{m+n-1}{k-1}. $