Show that $\frac{(m+n)!}{m!n!}$ is an integer whenever $m$ and $n$ are positive integers using Legendre's Theorem.
Hi everyone, I seen similar questions on this forum and none of them really talked about how to apply the Legendre's theorem to questions like the one above.
I get that $\frac{(m+n)!}{m!n!}$ = $\binom{m+n}{m}$, which is an integer. But could someone explain how the Legendre proof works in this case and why it proves the above is true?
For a prime $p$, denote by $v_p(r)$ the exponent of $p$ in the prime factorization of $r$. So for example, $v_2(12) = 2$. Legendre's theorem states that for any prime $p$ and integer $n$, we have
$$v_p(n!) = \sum_{i = 1}^\infty \left\lfloor \frac{n}{p^i} \right\rfloor $$
Note that $v_p(rs) = v_p(r) + v_p(s)$. Also $r$ divides $s$ if and only if $v_p(r) \leq v_p(s)$ for all primes $p$.
The fraction $\frac{(m+n)!}{m!n!}$ is an integer if and only if $m!n!$ divides $(m+n)!$. So to prove your statement, it suffices to prove $v_p(n!) + v_p(m!) \leq v_p((n+m)!)$. Using Legendre's theorem, this follows from $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x + y \rfloor$.
