## Thinking Mathematically (6th Edition)

We have 5 different people, and we need to select 2 speakers for an event. For convenience, we'll refer to those people as A, B, C, D and E. If we select A, we can select either B, C, D or E to go as well, which means that there are 4 ways to select the members so that A goes. If we don't select A, and we select B, we can select C, D or E to go as well, which means that there are 3 ways to select the members so that A doesn't go and B goes. If neither A nor B goes, we can pair C and D, C and E or D and E, giving us the last 3 ways. This gives us a total of $4+3+3=10$ ways to select the speakers.