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 $n>0$?

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*}
