TITLE: Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.
QUESTION [6 upvotes]: This is also a question on my exam paper that i proved by using mathematical induction.
However, my tutor tells me that it can be proved without using mathematical induction.
I really want to know how to deal with in another way.
Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.

REPLY [12 votes]: Hint:
$$
\begin{align}
&\binom{3n}{3}\binom{3n-3}{3}\binom{3n-6}{3}\cdots\binom{6}{3}\binom{3}{3}\\
&=\frac{(3n)!}{\color{#C00000}{(3n-3)!}3!}\frac{\color{#C00000}{(3n-3)!}}{\color{#00A000}{(3n-6)!}3!}\frac{\color{#00A000}{(3n-6)!}}{\color{#A0A0A0}{(3n-9)!}3!}
\cdots\frac{\color{#A0A0A0}{6!}}{\color{#0000FF}{3!}3!}\frac{\color{#0000FF}{3!}}{0!3!}\\
&=\frac{(3n)!}{(3!)^n}
\end{align}
$$
Only the solid black terms don't disappear.