Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$ I'm trying to prove that : 
$$\frac{100!}{50!\cdot2^{50}}$$
is an integer .
For the moment I did the following : 
$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$
But it still doesn't quite work out . 
Hints anyone ? 
Thanks 

Combinatorial argument:
The number of ways to arrange the digits $1,1,2,2,3,3,4,4,5,5,\ldots, 50,50$ in a row is $\frac{100!}{2^{50}}$.
This is clearly a multiple of $50!$, since we can perform any of the $50!$ permutations on the set of 50 elements.
