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

Question

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 $6!$.

Answer 1

$$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}$$

$$\binom{3n}{n} = \binom{3n -1}{n-1} +\binom{3n-1}{n} = \frac{(3n - 1)!}{(n-1)!(3n -1 -(n-1))!} + \frac{(3n - 1)!}{n! ((3n -1 -n)!}$$ $$=\frac{(3n-1)!)}{(n-1)!(2n-1)!}(\frac{1}{2n} + \frac{1}{n})$$ $$=\frac{(3n-1)!)}{(n-1)!(2n-1)!} \frac{3}{2n}$$ $$=\frac{(3n-1)!)}{(n-1)!(2n)!} * 3$$ $$=3\binom{3n -1}{n-1}$$

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

$$\binom{3n}{n,n,n} = \binom{3n}{n}\binom{2n}{n} = 6\binom{3n -1}{n-1}\binom{2n-1}{n-1}$$

Answer 2

NOTE: This proof assumes $n-2 \geq 0$. The proof is trivial for $n=0$ and $n=1$.

By Pascal's Identity, we have the following: $${3n \choose n, n, n}={3n-1 \choose n-1, n, n}+{3n-1 \choose n, n-1, n}+{3n-1 \choose n, n, n-1}=3{3n-1 \choose n-1, n, n}$$

Use Pascal's Identity on the new binomial coefficient: $${3n-1 \choose n-1, n, n}={3n-2 \choose n-2, n, n}+{3n-2 \choose n-1, n-1, n}+{3n-2 \choose n-1, n, n-1}={3n-2 \choose n-2, n, n}+2{3n-2 \choose n-1, n-1, n}$$

In order to prove ${3n-2 \choose n-2, n, n}$ is divisible by $2$, we will now prove that ${2n+a \choose a, n, n}$ is divisible by $2$ by induction.

Base Case $a=0$: $${2n \choose 0, n, n}={2n \choose n}$$

This is divisible by $2$ by your proof.

Induction Case: $${2n+a \choose a, n, n}={2n+a-1 \choose a-1, n, n}+{2n+a-1 \choose a, n-1, n}+{2n+a-1 \choose a, n, n-1}={2n+a-1 \choose a-1, n, n}+2{2n+a-1 \choose a, n-1, n}$$

The former addend in this sum is divisible by $2$ by our induction hypothesis and the latter addend is obviously divisible by $2$, so the whole sum is divisible by $2$.

Thus, from this, we know that ${3n-2 \choose n-2, n, n}$ is divisible by $2$. Go back to our second sum: $${3n-1 \choose n-1, n, n}={3n-2 \choose n-2, n, n}+2{3n-2 \choose n-1, n-1, n}$$

The former addend in this sum is divisible by $2$ by our lemma and the latter addend is obviously divisible by $2$, so the whole sum is divisible by $2$.

Now, go back to our original equation: $${3n \choose n, n, n}=3{3n-1 \choose n-1, n, n}$$

A multiple of $2$ times $3$ is obviously divisible by $6$, concluding the proof.

Answer 3

Just count in how many ways you can partition a set of $3n$ elements into three sets of $n$ elements ignoring order.

If we consider the order, we get $\binom{3n}{n,n,n}$ ordered partitions. Each unordered partition is counted exactly $3!$ times (do you see why?), so the number of unordered partitions is exactly

$$\frac{1}{3!}\binom{3n}{n,n,n}$$

and it is necessarily an integer.