How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$? How can I prove that  $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$?
 A: 
If you have $3$ numbered buckets, and $3n$ items, then there are $\binom{3n}{n,n,n}$ ways to put $n$ in each bucket.The buckets are identical apart from the fact that we've numbered them. If someone had secretly switched their labels around, there would still be $\binom{3n}{n,n,n}$ ways to have $n$ in each.If we remove the labels from the buckets however, we now consider any permutation of the buckets to be identical. Each of those 3! permutations has the same number of arrangements for the balls, and we know the total number of those arrangements is $\binom{3n}{n,n,n}$.Therefore, the number of ways to arrange $3n$ balls in $3$ identical buckets with $n$ balls each is $\binom{3n}{n,n,n}/3!$, and this must obviously therefore be an integer.

A: Referring to the standard result, the number of factors of $p$ in $\frac{(3n)!}{n!n!n!}$ is given by
$$
\sum_{i \ge 1} \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \sum_{i \ge 1} \left\lfloor \frac{n}{p^i} \right\rfloor
= \sum_{i \ge 1} \left( \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \left\lfloor \frac{n}{p^i} \right\rfloor \right).
$$
All of the terms of the sum are nonnegative. We have to prove that when $p =2$ or $3$, there is a term that is strictly positive.


*

*For $p = 3$, let $3^k$ be the highest power of $3$ dividing $n$. Then the term $i = 3^{k+1}$ works:
$$
\left\lfloor \frac{3n}{3^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{3^{k+1}} \right\rfloor = 3 \left[\frac{n}{3^{k+1}} -\left\lfloor \frac{n}{3^{k+1}} \right\rfloor \right] > 0.
$$

*For $p = 2$, let $2^k$ be the highest power of $2$ dividing $n$, and again take the term $i = 2^{k+1}$:
\begin{align*}\left\lfloor \frac{3n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor
&= \left\lfloor \frac{2n}{2^{k+1}} + \frac{n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor \\
&= \frac{2n}{2^{k+1}} + \left\lfloor \frac{n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor \\
&= 2 \left[\frac{n}{2^{k+1}} - \left\lfloor \frac{n}{2^{k+1}} \right\rfloor\right] \\
&> 0.
\end{align*}
