Let $X_n$ be the set of all word of the length $3 n$ over the alphabet $\{A,B,C\}$ which contain each of the three letters n times.
The amount of elements of $X_n$ is $\frac{(3n)!}{(n!)^3}$, but why?
I tried to split the problem into three smaller ones by only looking at the distribution of the single letters. For each one there should be $\binom{3n}{n} = \frac{(3n)!}{n!(2n)!}$ possibilities, but if this is correct, how do I get it combined? According to Wolfram Alpha $\binom{3n}{n}^3 \neq \frac{(3n)!}{(n!)^3}$.
Thanks in advance!