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 $6!$. 

Notice that $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}.$$ We have shown that $\binom{2n}{n}$ is divisible by 2. Now, all we must do is show that $\binom{3n}{n}$ is divisible by 3. You know that $\binom{3n}{n}$ is the number of n-element subsets of a 3n-element set (or, the number of ways to choose n objects among 3n distinct objects), which is always an integer. Note that $$\binom{3n}{n} = \frac{(3n)!}{(n!)(2n)!} = \frac{(3)(n)(3n - 1)!}{(n)(n-1)!(2n)!} = \frac{(3)(3n - 1)!}{(n-1)!(2n)!} = (3)\frac{(3n - 1)!}{(n-1)!(2n)!} = (3)\binom{3n-1}{n-1},$$ which is an integer and is divisible by 3.
By Peter's suggestion, you can generalize to say that for all integers k, $$\binom{kn}{n_{1},n_{2}, ... , n_{k}} = \binom{kn}{n}\binom{(k-1)n}{n}...\binom{(k-k+1)n}{n}$$ and $$\binom{kn}{n} = \frac{(kn!)}{n!(kn-n)!} = \frac{(kn)(kn-1)!}{(n)(n-1)!(kn-n)!} = k\binom{kn-1}{n-1}, $$ and use induction to prove that $$\binom{kn}{n_{1},n_{2}, ... , n_{k}}$$ is divisible by $k!.$
