Prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer. 


  For clarity: the denominator is the only part being squared.


My thought process: 

The numerator is the product of the first $n$ even numbers and the product of the first $n$ odd numbers. 

That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with $n!$ resulting in the following quotient: 
$$ \frac{(2^n)(2n-1)(2n-3)}{(n!)}\;.$$ 
To me this looks even thanks to the powers of $2$. But I am not convinced for some reason. 

Was this approach to naive? 

Sorry for the poor notation, I don't know any coding languages, my apologies. 

We have $$\displaystyle \frac{(2n)!}{(n!)^2} = \binom{2n}{n} $$ and of course for every way  to choose a combination of $n$ objects from a total of $2n$ objects, there exists a complementary choice (the left over $n$ objects). Since the ways to pick $n$ objects from $2n$ comes in pairs, it follows that the total number is even. 

Another proof: $$ \frac{(2n)!}{(n!)^2} = \frac{ 2n \cdot (2n-1)! }{(n!)^2 } = 2 \cdot \frac{ n \cdot (2n-1)! }{ n\cdot (n-1)! \cdot n!} = 2 \cdot \frac{(2n-1)!}{n! (n-1)!} = 2 \binom{2n-1}{n} $$

This one can be written in another flavor (using $ \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ and the symmetric property):  $$ \frac{(2n)!}{(n!)^2} = \binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n} = 2\binom{2n-1}{n}.$$
Yet another: The sum of the $2n$-th row of Pascal's triangle is $2^{2n}$ which is even, and the sum of all the entries excluding the central coefficient is also even, because of the symmetric property $ \displaystyle \binom{n}{k} = \binom{n}{n-k} $. Thus, the central coefficient must also be even.



We can squeeze this problem a bit further and learn something interesting with this fourth proof. This one does not require any knowledge of binomial coefficients. To investigate the prime factors of factorials we use the following identity:  $$ l = \sum_{k=1}^{\infty} \bigg\lfloor \frac{n}{p^k} \bigg\rfloor $$

where $p$ is a prime number and $l$ is the unique natural number such that $p^l \mid n!$ but $p^{l+1} \nmid n!.$ Basically this counts the multiplicity of the prime factor $p$ in $n!$. 

The idea of the proof is this: Write $n! = 1\cdot 2 \cdot 3 \cdots (n-1) \cdot n$, and ask yourself where the factors of $p$ come from. Clearly you get one factor of $p$ from $p, 2p, 3p, \cdots $ so there are $ \lfloor n/p \rfloor $ factors of $p$ from these. But that's not the whole story - when we checked over the multiples $p,2p,3p\cdots $ we didn't count the second factor of $p$ every time there was $p^2, 2p^2, 3p^2 \cdots $ so there comes another $ \lfloor n/p^2 \rfloor $ to add. Wait, we didn't count another factor every time we missed cubes in $p^3, 2p^3,\cdots$, so we add another $ \lfloor n/p^3 \rfloor $ and so on. Also, note that this may look like an infinite series, but eventually $ p^k > n$ so all the terms eventually become $0. $

Anyway, back to the main goal. Using our identity, we see that there are $$ \sum_{k=1}^{\infty} \bigg \lfloor \frac{2n}{2^k} \bigg \rfloor - 2 \bigg \lfloor \frac{n}{2^k} \bigg \rfloor$$ factors of $2$ in $ \displaystyle \frac{(2n)!}{ (n!)^2} $, and we wish to show that this number is at least $1.$ 

With some easy casework we can see that $$
    \lfloor 2x \rfloor - 2 \lfloor x \rfloor  = \left\{
     \begin{array}{lr}
       1 & : \{ x\} \geq \frac{1}{2} \\
       0 & : \{ x\} < \frac{1}{2} 
     \end{array}
   \right.
$$ where $\{ x \} $ denotes the fractional part of $x.$ Thus our problem is reduced to showing that there exists some $ k\in \mathbb{N} $ such that $ \displaystyle \frac{n}{2^k} $ has fractional part greater than or equal to $1/2.$ This is of course true, since if $m$ is the largest positive integer such that $\displaystyle \frac{n}{2^m} \geq 1 $ then $ \displaystyle \frac{1}{2} \leq \frac{n}{2^{m+1} } < 1.$ Hence, $ \displaystyle \frac{(2n)!}{(n!)^2} $ is even. 
