8
$\begingroup$

I know a proof that $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ is an integer.

The proof goes as:

If a prime $p$ divides $(15n)!(10n)!(6n)!$, then the power of the prime dividing $(15n)!(10n)!(6n)!$ is lesser than the $(30n)!n!$. It is relatively easy to prove this.

$\textbf{My question is there a counting argument to prove that this is an integer?}$

By this, I mean does $\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$ count something?

(I have a gut feeling that this should count something but I have not thought in depth about this).

For instance $\frac{n!}{r!(n-r)!}$ counts the number of ways of choosing $r$ objects out of $n$.

Also, are there other examples similar to this? (I tried searching for other such examples in vain)

  • 0
    @Sivaram: I think I misremembered the second example. See my answer.2010-11-12

1 Answers 1

4

This MO thread has a lot of good information. I think we should be pessimistic. Pietro Majer's answer and the comments therein suggest that this would be hard, at least in the sense that one should not expect a proof anything like the proof for binomial coefficients.