TITLE: How many ways are there to split $10$ people into two groups of $5$?
QUESTION [14 upvotes]: In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $\binom{10}{5}$. 
In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?
Example
There are ten people in a basketball squad. Find how many ways:

the starting five can be chosen from the squad
the squad can be split into two teams of five.

Solution

There are $\binom{10}{5} = \frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1} = 252$ ways of chosing the starting five.
The number of ways of dividing the squad into two teams of five is $\frac{252}{2} = 126$.

REPLY [3 votes]: Imagine that you're one of the $10$ players.  In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players.  This can be done in $\binom{9}{4} = 126$ ways.
Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets.  The two outcomes are rather different for you, so we have to count them differently.  There are thus $2\binom{9}{4} = 252$ possible outcomes.
On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently.  It is not really possible to distinguish between the following cases:
(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players
or 
(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.
Both (I) and (II) describe the same event.  That event should not be counted twice.  So there are just $\binom{9}{4} = 126$ possible outcomes here.