Prove by counting in two different ways I'm am not completely clear on what is means to count in two different ways. Here is the question. $m$ and $n$ are integers with $0 \le m \le n$. In a town in the USA there are $n+1$ townspeople. This town is run by one mayor and $m$ council-members (the mayor cannot be a councillor). Prove the following by counting in two different ways the number of ways to choose a town's elected officials. $(n+1)$$n \choose m$$=(n+1-m)$${n+1}\choose m$ So if anyone can explain how to do these types of questions that would be great. Thanks

These kinds of proofs are called **combinatorial proofs**. In order to choose a mayor and $m$ council-members, we can do it in two ways: we can choose a mayor and then the council-members, or we can choose the council-members and then choose the mayor.

In the first way, we have $n + 1$ choices for the mayor. To choose a council, we need to pick $m$ people from the remaining $n$ townspeople (as the mayor cannot be a councillor). Thus, there are $\binom{n}{m}$ ways to choose the council; multiplying the two together gives $(n + 1)\binom{n}{m}$ ways in total.

Can you see how to get the other formula combinatorially?