<s>[INST] How does one prove that $6$ divides ${(3n)!}/{(n!)^3}$? How does one prove that $6$ divides the multinomial coefficient $\displaystyle \frac{(3n)!}{\;(n!)^3}$?
 [/INST] $\displaystyle
\frac{(3n)!}{\;(n!)^3} = 3n\frac{(3n-1)\cdots(2n)}{n!} \, \frac{(2n-1)\cdots n}{n!} \, \frac{(n-1)\cdots 1}{n!} =  3 {3n-1 \choose n} {2n-1 \choose n} 
$
$\displaystyle
\frac{(3n)!}{\;(n!)^3} = \frac{(3n)\cdots(2n+1)}{n!} \, 2n \, \frac{(2n-1)\cdots n}{n!} \, \frac{(n-1)\cdots 1}{n!} =  2 {3n \choose n} {2n-1 \choose n} 
$
So
$\displaystyle
\frac{(3n)!}{\;(n!)^3}$ is a multiple of $2$ and of $3$, hence a multiple of $6$.
Here is a simpler solution:
$\displaystyle
\frac{(3n)!}{\;(n!)^3} = {3n \choose n} {2n \choose n} = \frac{3n}{n}{3n-1 \choose n-1} \frac{2n}{n} {2n-1 \choose n-1} = 6 {3n-1 \choose n-1} {2n-1 \choose n-1}
$
This solution can be generalized to prove that $\displaystyle
\frac{(kn)!}{\;(n!)^k}$ is a multiple of $k!$.
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