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

Solution 1:

$$ \frac{(2n)!}{n! 2^{n}} = \frac{\prod\limits_{k=1}^{2n} k}{\prod\limits_{k=1}^{n} (2k)} = \prod_{k=1}^{n} (2k-1) \in \Bbb{Z}. $$

Solution 2:

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.