divisibility of factorials: $a!b!(a+b)! \mid (2a)!(2b)!$

Suppose $a,b \in \mathbb{N}$, prove that $a!b!(a+b)! \mid (2a)!(2b)!.$

The only proof I know is to uses the legendre theorem

I wanna know if there is a one line proof like the above hyperlink or alternative way, thank you so much!

This is IMO $1972$'s $3$rd question.

We need to prove that $\dfrac{(2a)!(2b)!}{a!b!(a+b)!}$ is an integer.

Call the the given expression $f(a,b)=\dfrac{(2a)!(2b)!}{a!b!(a+b)!}$

Now notice that $f(a,b)=4f(m,n-1)-f(m+1,n-1) $

So, you have a recursion in which $n$ can be reduced until you get $0$.

$n$ can be reduced until one obtains $f(a,b)=\sum_ic_rf(r,0)$

Now $f(r,0)$ is a simple binomial coefficient, $c_r$'s are natural numbers.

The question was posed just to indicate not all such expressions can be expressed as a counting problem.

And it also has nice and simple proof involving Legendre theorem(As you have mentioned). Do post it as an alternate answer.:)

We have (see remark of dtldarek): $$ \frac{(2a)!(2b)!}{a!b!(a+b)!}=(-1)^a 2^{2a+2b}\binom{a-1/2}{a+b}. $$ This is integral, because it appears among the coefficients of odd powers of
$$ (1-4z)^{-1/2}=\sum_{n=0}^{\infty}\binom{2n}{n}z^n, $$ which has integral coefficients. More precisely, $$ (1-4z)^{a-1/2}=\sum_{b=0}^{a-1}(-1)^b\frac{(a-b)!(2a)!}{(2a-2b)!a!b!}z^b+(-1)^a\sum_{b=0}^{\infty}\frac{(2a)!(2b)!}{a!b!(a+b)!}z^{a+b}. $$