# Counting ways to divide $10$ kids into $2$ teams of $5$. Why divide by $2!\,$?

Here's the question:

In order to play a game of basketball, $$10$$ kids at a playground divide themselves into two teams of $$5$$ each. How many different divisions are possible?

The solution given is:

$$\frac{10!}{5!5!2!}$$

My question is

Why is the answer not just $$\frac{10!}{5!5!}$$ i.e, Why did they divide by $$2!$$?

Let's line up all ten people, and let's say that the first five go on team 1, and the next five on team 2. Now, there are $$10!$$ different ways to line up the $$10$$ people. However, note that you get the same two teams if:
1. you shuffle the first five people. Having, say, persons 2 and 4 change positions in the line-up will not change the teams: persons 2 and 4 will still be on team 1. Since there are $$5!$$ ways to shuffle the first five people, you need to divide the original $$10!$$ by $$5!$$
2. you shuffle the persons in positions 6 through 10. Same story. So again, divide by $$5!$$
3. You swap the first five people and the second five people. If the teams are indistinct, then the same five people will be on one team, and the same other five on another team. So: divide by $$2$$ or, what is the same thing: divide by the number of ways we can shuffle the groups, and since there are two groups, that can be done in $$2!$$ ways. Of course, if the teams are distinct (e.g if the first team is 'the Flyers' and the second 'the Eagles', then swapping the groups of people will make a difference, so in that case do not divide by $$2!$$