TITLE: Using combinatorial argument prove that $\frac{(3n)!}{2^n\times 3^n}$ is an integer.
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: Since $2^n$ and $3^n$ are always coprime, you can split up the problem into two parts:

$3^n \vert (3n)!$ (You already did that yourself.)
$2^n \vert (3n)!$ (But analogously to your own consideration, one immediately sees that $2^n $ divides even $(2n)!$, which is a divisor of $(3n)!$

This proves $6^n=\text{lcm}(2^n,3^n) \vert (3n)!$.