Examples 5b and 5c from "A First Course in Probability, 7th edition" by Sheldon Ross. Images of the relevant pages are attached. EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible? Solution. There are 10!/(5!.5!) = 252 possible divisions. EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible? Solution. Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is 10!/(5!.5!.2!) = 126 I disagree. I think for Example 5b the answer should be (10!/(5!.5!))*(10!/(5!.5!)). Because the teams A and B play in two different leagues, the same person can play in both teams. So there are 10C5 = 10!/(5!.5!) ways of choosing 5 people for team A, and the same number of ways for choosing 5 people for team B. For example 5c, I think the answer should be 10C5 = 10!/(5!.5!). The two teams seem to be playing against each other, so no one can play both teams. So the number of ways of choosing the teams is just the number of way of choosing 5 people from 10. The remaining 5 will make up the other team. But maybe the author of the book and I are not thinking along the same lines. Can someone point out my error and show me where I went wrong? Thanks.