TITLE: Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$
QUESTION [31 upvotes]: 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

REPLY [8 votes]: $$100!=(1\cdot 3\cdot 5\cdots 99)\cdot (2\cdot 4\cdot 6\cdots 100),$$
where
$$(2\cdot 4\cdot 6\cdots 100)=2^{50}\cdot (1\cdot 2\cdot 3\cdots 50)=2^{50}\cdot 50!$$
so
$$(50\cdot 2)! = 2^{50}\cdot 50!\cdot (1\cdot 3\cdot 5\cdots 99)$$

More generally 
$$(m\cdot n)! \, =\,  m^{n}\  \cdot\  n!\ \cdot\ \prod_{(k,\, j) \ =\ (1,\,1)}^{(n,\, m-1)}\,  (m\cdot k-j)$$
Your case is $n=50, m=2$ and the product on the right is a natural number.