Can anyone help get me started on this?
Provide a combinatorial argument to show that if n and k are positive integers with n = 3k, then n!/(3!)^k is an integer.
The second step states to "generalize" the result of the above.
Hello TulkiA well-known result states that the number of arrangements of $\displaystyle n$ items, where $\displaystyle n_1$ of the items are identical of the first kind, $\displaystyle n_2$ items are identical of the second kind, ... is $\displaystyle \frac{n!}{n_1!n_2!...}$. See, for example, just here.
So if we have $\displaystyle n=3k$ items, which can be sorted into $\displaystyle k$ groups of $\displaystyle 3$, each group containing $\displaystyle 3$ identical items, then the number of arrangements of the $\displaystyle n$ items is $\displaystyle \frac{n!}{(3!)^k}$. This number is therefore an integer.
To generalise this, the same argument applies if $\displaystyle n = mk$, for any positive integer $\displaystyle m$, whereby $\displaystyle \frac{n!}{(m!)^k}$ is therefore also an integer.
Grandad