Number of ways to divide n boys onto two football teams of $\frac{n}{2}$ each? Here n is even
Attempt: Is it just $n \choose \frac{n}{2} $. I am getting confused whether should I divide by 2 or not. Is this question any different from having 20 chocolates and putting them in 2 bags so that each contains 10 chocolates. Is this just a generalization of this?
If the teams are labeled, then the answer is $$\binom{n}{\frac{n}{2}}$$ However, if they are not labeled, the answer is $$\frac{1}{2}\binom{n}{\frac{n}{2}}$$ The reason is that when you select a particular group of boys to form a team, you are also choosing the opposing team. Therefore, you are selecting the same two teams twice, once when you select a particular group of boys to be on a team and once when you choose the opponents of that group of boys to be on a team.
If the chocolates are distinguishable, then the answer is analogous to the problem above. If the bags are distinguishable, there are $$\binom{20}{10}$$ ways of distributing the chocolates. If the bags are indistinguishable, there are half that number of ways to distribute the chocolates.