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!
 A: 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.:)  
A: 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,
$$
(1-4z)^{a-1/2}=\sum_{b=0}^{a-1}(-1)^b\frac{(a-b)!(2a)!}{(2a-2b)!a!b!}z^b+(-1)^a\sum_{b=0}^{\infty}\frac{(2a)!(2b)!}{a!b!(a+b)!}z^{a+b}.
$$
