Three teams of four players

Text

In how many ways can $12$ distinct players be divided into three teams of $4$, labelled $A$, $B$ and $C$?

Solution

First we choose the $4$ players for team $A$, then $4$ for $B$ among those left, and the last $4$ form $C$: $N = \binom{12}{4}\binom{8}{4}\binom{4}{4} = 495\cdot 70\cdot 1.$ Equivalently, it is the multinomial coefficient $N = \frac{12!}{4!\,4!\,4!} = \frac{479\,001\,600}{24^{3}} = \frac{479\,001\,600}{13\,824} = \boxed{34\,650}.$ (The teams are labelled, so we do not divide by $3!$.)