1. ## Combinatorial Argument

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.

2. Hello Tulki
Originally Posted by Tulki
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.
A well-known result states that the number of arrangements of $n$ items, where $n_1$ of the items are identical of the first kind, $n_2$ items are identical of the second kind, ... is $\frac{n!}{n_1!n_2!...}$. See, for example, just here.

So if we have $n=3k$ items, which can be sorted into $k$ groups of $3$, each group containing $3$ identical items, then the number of arrangements of the $n$ items is $\frac{n!}{(3!)^k}$. This number is therefore an integer.

To generalise this, the same argument applies if $n = mk$, for any positive integer $m$, whereby $\frac{n!}{(m!)^k}$ is therefore also an integer.

3. ## Thanks!

Luckily, we just had a lecture today which covered this concept, so your contribution makes sense.