How many ways are there of splitting twelve people into two groups of the same size? Twelve people need to be split up into teams for a quiz. How many ways are there of splitting them into two groups of the same size?
I did $12 C 6$, which gives $924$, however the answer is $\frac{1}{2}(12 C 6)$, which is $462$. Why is the $12 C 6$ halved?
Any help will be much appreciated, thanks in advance. 
 A: Actually, it depends on whether the teams are labelled or unlabelled.
Your answer is for labelled teams whereas the book answer is for  unlabelled teams.
For instance, you can divide 1234 as 12|34, 13|24, 14|23, 23|14, 24|13 and 34|13.
But if they are unlabelled teams, each pair like 12|34 and 34|12 are the same.
The question doesn't say labelled teams (e.g. team A and team B), so we take it as unlabelled teams. 
A: $12\choose 6$ double counts.  For instance you could choose A, B, C, D, E, F for one group, leaving G, H, I, J, K, L for the other; or you could choose G, H, I, J, K, L for one group, leaving A, B, C, D, E. F.  But this actually gives you the same two groupings.
A: Think about it as such: first of all put these people in one straight line , you can rearrange them in $12!$ ways. The first six from the left to the right is one team, and the other 6 the second one ( you don't have to worry about that). Now you have to fix the fact that you don't want to count multiple times for things like A B C D X Y | E F G H Q P and B A C D X Y | E F G H Q P (i just changed the order of people A and B) . Once, you compute a possible team as (A ,B ,C,D,X,Y) (E , F , G ,H, Q ,P) make sure you exclude their rearrangements ($6!$ for each). Finally , make sure you don't count twice facts like :E F G H Q P | A B C D X Y (we paired the exact same people ) so divide by $2!$ to fix that . Generally , if you need to separate $n$  people into $r$ teams with $n_1, n_2, .., n_r$ people each $\frac{n!}{n_1!n_2!...n_r}$. As mentioned above and as stated in your problem : if teams are not "labeled" - we only care which people pair with which- then those "created " teams have $r!$ rearrangements so divide by this and $\frac{n!}{n_1!n_2!...n_r*r!}$ and you are done.
