There are 3n different cells, 2n white balls, and n black balls.
Each cell cannot be empty and must contain exactly one ball. How many ways can we put those balls into those cells?
My solution is:
Let's start with putting all the 2n white balls in the cells $\binom{3n}{n}$.
than I have n cells and n black balls $\binom{n}{n}$.
we get: $\binom{3n}{n}$ * $\binom{n}{n}$.
Is this correct?