Show that (2a)!(2b)!/(a!b!(a + b)!) is an integer.
Rearranging, you get $\displaystyle \frac{(2a)!}{a!}*\frac{(2b)!}{b!}*\frac{1}{(a+b)!}$
Which equals $\displaystyle \frac{(_{2a}P_a)*(_{2b}P_b)}{(a+b)!}$
In plain English, you are picking $\displaystyle a$ objects from $\displaystyle 2a$ options and $\displaystyle b$ objects from $\displaystyle 2b$ options with regard to order. Thus, the $\displaystyle (a+b)!$ term is simply dividing out all possible orderings of your $\displaystyle a+b$ objects.
I am sure someone can come up with a formal proof, but this at least should be convincing