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

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.

Solutions Collecting From Web of "prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer"

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 $$
\lfloor 2x \rfloor – 2 \lfloor x \rfloor = \left\{
1 & : \{ x\} \geq \frac{1}{2} \\
0 & : \{ x\} < \frac{1}{2}
$$ 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.

It is often fun to try to solve such problems using Lagrange’s Theorem from finite group theory. That is, to prove that $a$ divides $b$ you try to find a group of cardinality $b$ with a subgroup of cardinality $a$.

In this case, the symmetric group on $2n$ letters, $S_{2n}$, is an obvious choice for a group of cardinality $(2n)!$. Let $G\subset S_{2n}$ be the set of all permutations which map the set $\{1,\ldots, n\}$ onto $\{1,\ldots, n\}$ or $\{n+1,\ldots, 2n\}$. You can check that $G$ is a subgroup of $S_{2n}$.

To compute the cardinality $\lvert G \rvert$, one option is to observe that $G\cong S_n\wr \mathbb{Z}_2$, so $\lvert G\rvert = \lvert S_n\rvert^{\lvert \mathbb{Z}_2\rvert} \lvert \mathbb{Z}_2\rvert = 2(n!)^2$. Alternatively, choosing an element of $G$ corresponds to choosing two permutations of a set of size $n$ and a choice of whether to swap $\{1,\ldots,n\}$ with $\{n+1,\ldots, 2n\}$ or not, so $\lvert G\rvert = 2(n!)^2$.

Thus by Lagrange’s Theorem $2(n!)^2$ divides $(2n)!$, or in other words, $\frac{(2n)!}{(n!)^2}$ is even.

By the binomial theorem,
4^n=(1+1)^{2n}=\sum\limits_{k=0}^{2n}{2n\choose k}={2n\choose n}+\sum\limits_{k=0}^{n-1}{2n\choose k}+\sum\limits_{k=n+1}^{2n}{2n\choose k}.
Since Pascal’s triangle is symmetrical, each term in the sum from $k=0$ to $n-1$ coincides with one term in the sum from $k=n+1$ to $2n$, hence these two sums are equal and
\frac{(2n)!}{(n!)^2}={2n\choose n}=4^n-2\cdot\sum\limits_{k=0}^{n-1}{2n\choose k}
is even for every $n\geqslant1$.

This answer has been moved from this question, which was closed because it was judged to be a duplicate of this question.

The number of factors of a prime $p$ in $n!$ is
$\left\lfloor\frac{n}{p}\right\rfloor$ counts the number of positive multiples of $p$ no greater than $n$, but only once. $\left\lfloor\frac{n}{p^2}\right\rfloor$ counts the multiples of $p^2$ a second time, and $\left\lfloor\frac{n}{p^3}\right\rfloor$ the positive multiples of $p^3$ a third time, etc. Applying $(1)$ to the base-$p$ representation of $n$
where $0\le n_i<p$, yields
where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.

Therefore, the number of factors $p$ in $\displaystyle\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is
Using $(4)$ with $p=2$ says that the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is $\sigma_2(n)+\sigma_2(n)-\sigma_2(2n)$ which is simply $\sigma_2(n)$ since $\sigma_2(n)=\sigma_2(2n)$. Therefore, the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is the number of $1$-bits in the binary representation of $n$.

Thus, if $n>0$, then there is at least one $1$-bit in $n$ in binary, and so $2$ divides $\displaystyle\binom{2n}{n}$.

Let’s observe an example:

$(2\cdot 5)!=10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!$ , we may conclude that:

$(2n)!=2n\cdot(2n-1)\cdot(2n-2)….(n+1)n!$ so we get following expression :

$$\frac{2n\cdot(2n-1)\cdot(2n-2)….(n+1)}{n!}=\frac{2n\cdot(2n-1)\cdot2\cdot(n-1)\cdot(2n-3)\cdot 2 \cdot (n-2)….(n+1)}{n!}$$

$$=\frac{2\cdot2\cdot2…\cdot2\cdot n\cdot(n-1)\cdot(n-2)\cdot……\cdot 1\cdot(2n-1)\cdot(2n-3)\cdot……(n+1)}{n!}$$

$$=\frac{2\cdot2\cdot2…\cdot2 \cdot n!\cdot(2n-1)\cdot(2n-3)\cdot……\cdot (n+1)}{n!}=$$

$$=2\cdot2\cdot2…\cdot2 \cdot(2n-1)\cdot(2n-3)\cdot……\cdot (n+1)$$ so number is even.

The exact power of $2$ dividing $\binom{2n}{n}$ is $2$^(sum of digits in base $2$ representation of $n$).

The exact power of $3$ dividing the trinomial coefficient $\binom{3n}{n,n,n} = (3n)!/(n!)^3$ is $3$^(sum of digits in base $3$ representation of $n$).

The exact power of $p$ dividing the $p$-nomial coefficient $\binom{pn}{n,n,\dots , n} = (pn)!/(n!)^p$ is $p$^(sum of digits in base $p$ representation of $n$) for any prime $p$.

This is from the formula for the power of $p$ dividing $n!$, which is ($n$ – sum of digits of $n$ when written in base $p$)/($p-1$). There also are combinatorial and generating function proofs but they are more complicated.