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 
 A: $$100!=(1\cdot 3\cdot 5\cdots 99)\cdot (2\cdot 4\cdot 6\cdots 100),$$
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.
A: Prove $51\cdot 52\cdots 100$ is divisible by $2^{50}$. Hint: Count how many multiples of 2, 4, 8... there are in the former expression.
A: From here and here, we have that the highest power of a prime $p$ dividing $n!$ is given by
$$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{p^k} \right\rfloor$$
Hence, the highest power of a prime $p$ dividing $\dfrac{n!}{m!}$ is given by
$$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{p^k} \right\rfloor - \sum_{k=1}^{\infty}\left\lfloor\dfrac{m}{p^k}\right\rfloor$$
I trust you can finish it off from here.
A: 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.
A: for the number of multiples of $2$:
$$100 = 52 + 2(n_1 - 1)$$
$$n_1 = 25$$
for the number of multiples of $4$:
$$100 = 52 + 4(n_2 - 1)$$
$$n_2 = 13$$
for the number of multiples of $8$:
$$96 = 56 + 8(n_3 - 1)$$
$$n_3 = 6$$
for the number of multiples of $16$:
$$96 = 64 + 16(n_4 - 1)$$
$$n_4 = 3$$
for the number of multiples of $32$:
$$96 = 64 + 32(n_5 - 1)$$
$$n_5 = 2$$
for the number of multiples of $64$:
$$64 = 64 + 64(n_6 - 1)$$
$$n_6 = 1$$
To get the number of $2$'s as factors, add all the $n$'s up, which yields $50$.
This should cancel the $2^{50}$ at the denominator, and thus prove that the expression is an integer.
A: 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.
A: $$ \frac{(2n)!}{n! 2^{n}} = \frac{\prod\limits_{k=1}^{2n} k}{\prod\limits_{k=1}^{n} (2k)} = \prod_{k=1}^{n} (2k-1) \in \Bbb{Z}. $$
