Prove $\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$ is always divisible by $6$ when $n$ is an integer. 
Prove $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$$ is always divisible by $6$ when $n$ is an integer. 

I have done a similar proof that $\binom{2n}{n}$ is divisible by $2$ by showing that $$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}$$ but I am at a loss for how to translate this to divisible by $6$. Another way to do this proof would be to show that when you shoot an $n$-element subset from $2n$ you can always match it with another subset (namely the $n$-elements that were not chosen). Again, no idea how to translate this to $6!$. 

$$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}$$
$$\binom{3n}{n} = \binom{3n -1}{n-1} +\binom{3n-1}{n} = \frac{(3n - 1)!}{(n-1)!(3n -1 -(n-1))!} + \frac{(3n - 1)!}{n! ((3n -1 -n)!}$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!}(\frac{1}{2n} + \frac{1}{n})$$
$$=\frac{(3n-1)!)}{(n-1)!(2n-1)!} \frac{3}{2n}$$
$$=\frac{(3n-1)!)}{(n-1)!(2n)!} * 3$$
$$=3\binom{3n -1}{n-1}$$
It has been already proved that
$$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}$$
Combining both
$$\binom{3n}{n,n,n} = \binom{3n}{n}\binom{2n}{n} = 6\binom{3n -1}{n-1}\binom{2n-1}{n-1}$$
