I'm trying to get a combinatorial proof of the following identity by making up some story.
$$\sum_{k=1}^n {k \choose j}k = {n+1 \choose j+1}n - {n+1 \choose j+2}$$
I can do it, by simplifying the expression first, but I wanted to prove it without any changes to the identity.
Suppose we have $n+1$ people and we would like to select $j+1$ of them to be on a sports team, with one of these being captain. The $n+1$ people are ordered in increasing age and we have been told that the captain should be the older than anyone else on the team. Furthermore we need to pick someone to organise the fixtures, and this person may not necessarily be one of the $j+1$ on the team (but they must be younger than the captain).
If the $(k+1)$st person is captain we can select $\binom{k}{j}$ teams from this and we can pick any of the $k$ people younger than the captain to be the organiser, giving $\binom{k}{j} k$. Since the captain could be any one of the $n+1$ people (except the first $j$, which correspond to the $k$ for which $\binom{k}{j}=0$ anyway), this gives $\sum_{k=1}^n \binom{k}{j} k$ teams in total.
On the other hand, we could instead first choose our team of $j+1$ from the $n+1$ people and then select someone (from the $n$ people other than the captain) to be the organiser, which can be done in $\binom{n+1}{j+1}n$ ways. The problem now is that the organiser could be older than the captain so we need to subtract the number of ways this could happen. This is equivalent to just choosing $j+2$ people from the $n+1$ (so the oldest is the organiser and next oldest is the captain), so in total we have $\binom{n+1}{j+1}n-\binom{n+1}{j+2}$ teams. Thus we have shown $$\sum_{k=1}^n \binom{k}{j} k=\binom{n+1}{j+1}n-\binom{n+1}{j+2}.$$