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 

We have $100$ people at a dance class. How many ways are there to divide them into $50$ dance pairs of $2$ people each? (Of course we will pay no attention to gender.)
Clearly there is an integer number of ways. Let us count the ways.
We solve first a different problem. This is a tango class. How many ways are there to divide $100$ people into  dance pairs, one person to be called the leader and the other the follower? 
Line up the people. There are $100!$ ways to do this. Now go down the line, pairing $1$  and $2$ and calling $1$ the leader, pairing  $3$ and $4$ and calling $3$ the leader, and so on.
We obtain each leader-follower division in $50!$ ways, since the groups of $2$ can be permuted. So there are $\dfrac{100!}{50!}$ ways to divide the people into $50$ leader-follower pairs to dance the tango.
Now solve the original problem. To just count the number of democratic pairs, note that interchanging the leader/follower tags produces the same pair division. So each democratic pairing gives rise to $2^{50}$ leader/follower pairings. It follows that there are $\dfrac{100!}{2^{50}\cdot 50!}$ democratic pairings.
