In how many different ways can you split $10$ people into two groups of $5$ people? In how many different ways can you split $10$ people into two groups with the same amount of people?
My attempt:
Since the order in which you choose someone doesn't matter, I chose to calculate the combinations:
$10 \choose 5$
 A: If the groups are labeled, your answer is correct.  Otherwise, notice that choosing five people to be in the first group automatically determines who is in the second group.  When you count combinations of five of the ten people, you count each group twice, once when you select the group and once when you select its complement.  Therefore, unless the groups are labeled, the number of ways to split ten people into two groups of five is $$\frac{1}{2}\binom{10}{5}$$  
Edit:  @GlenO proposed the following alternative argument in the comments:  Suppose Alicia is one of the ten people.  The number of ways two groups of five people can be formed is the number of ways of selecting four of the remaining nine people to be in Alicia's group or, equivalently, the number of ways of selecting five of the remaining nine people to not be in Alicia's group.  Therefore, the number of ways to split two people into two groups of five is 
$$\binom{9}{4} = \binom{9}{5}$$
which is equivalent to the answer I obtained above.
A: Selecting one team of five is \binom{10}{5}, not \binom{10}{2}. Then, the other five form the other team. You probably don't care which set of five forms the first team, so need to divide by 2 because you have counted each group twice.
If we consider a smaller version of the problem let's say we wanna select 2 teams each consist of 2 so, (ABCD) has (AB) (AC) (AD) (BC) (BD) (DC) teams, we divide this teams into two, getting (AB+CD),(AC+BD), (AD+BC).
You can also think of it this way, the oldest student X among the ten shall choose 4 individuals. This can be done in 9 choose 4 which is 126 ways. This gives the team containing X. The remaining students form the other team.
