TITLE: In how many different ways can two teams of five players each be formed?
QUESTION [7 upvotes]: Precalculus textbook problem (self-study): Ten people wish to play in a basketball game. In how many different ways can two teams of five players each be formed?
Textbook section title: Distinguishable Permutations and Combinations
Textbook solution in back of book: $C(10,5) = 252$.
My solution: $252/2 = 126$.
My reasoning: $C(10,5) = 252$ gives the number of five-person teams that can be formed, but that's double how many ways you can pick two teams because once one team is picked, the other is decided.
For example, suppose I have four people, A,B,C, and D.  If I ask how many different ways can two teams of two players each be formed from these four people the answer is three, not six:
Two teams: AB and CD
Two teams: AC and BD
Two teams: AD and BC
Yes, there are six different teams, but only three ways to pick two teams.
Question: Am I nuts?

REPLY [8 votes]: You’re not nuts!  But what your insight is really revealing is that this question is ambiguous. If there is a team A and a team B, then the book is right. But if the teams are not to be distinguished, then you are right. So which is it? We don’t know. That’s why the question is ambiguous.
Still, in a real life situation, where 10 players on a basketball court decide to create two teams, I would say your answer is more appropriate.

REPLY [3 votes]: You are correct that in picking one set of five the other is then chosen. However, the question assumes the teams are distinct in that C and D being on team one (blue) is not the same as C and D being on team two (red).
In your understanding, since you consider the two teams equivalent, you are correct in dividing by 2, as you are picking two indistinguishable teams simultaneously.