Using combinatorial argument prove that $\frac{(3n)!}{2^n\times 3^n}$ is an integer.
If we arrange $3n$ objects where there are 3 objects of one kind, another 3 objects of second kind $\cdots$ and 3 objects of $n^{th}$ kind then the number of ways are $\frac{(3n)!}{3^n}$. But I can't figure out how to introduce $2^n$ in the denominator.