A simple exercise in combinations I'm interested in the number of ways I can divide sixteen players on to four teams (with the order of selection being irrelevant). I know this is super simple, but I just wanted to check that I wasn't butchering it. I believe I can calculate this as follows:
(16 choose 4) * (12 choose 4) * (8 choose 4) = Just over 63 million
Are there really 63 million distinct ways to divide 16 players into four teams? I'm astounded by that number, and I'm therefore questioning my reasoning. I know this is simple, and I'm sheepish about asking it, but, if it's so simple, it won't take a moment to answer. And I really appreciate your help.
Many thanks.
 A: No, it is not correct.  What you calculated is the number of ways to select $4$ people to wear the blue uniforms, $4$ people to wear the pink uniforms, $4$ to wear the purple uniforms, and $4$ to wear black. You have to divide by $4!$ to get the right count.
A: Yes, it is correct. There are 63 million ways of going about the selection.
If you want another way of doing it consider this approach which (to me) is more intuitive:
\begin{align}
C=\dfrac{16!}{(4!)^4}
\end{align}
There are 16! ways of choosing people. Think of it as 16! possible ways of making 16 playesr stand in a line. But this accounts for repetition (or in other words, it gives importance to order) : Bob Sam Harry is treated different from Harry Bob Sam. To remove such combinations, realize that there are 4! combinations within each possible team and 4 teams in total. It will give you the same answer. 
EDIT : To address the issue of identification of teams:
If the teams also don't matter, then you further divide by $4!$ to give you close to 2 million combinations.
A: There are $16!/(4!)^4$ ways to distribute the 16 players over four teams of size 4. Since the order of the selection is irrelevant, we still can shuffle the teams and get the final number of possibilities $$\frac{16!}{(4!)^5} = 2.627.625\text{.}$$
