Proof that $\frac{(2n)!}{2^n}$ is integer

Question

I am trying to prove that $\dfrac{(2n)!}{2^n}$ is integer. So I have tried it by induction, I have took $n=1$, for which we would have $2/2=1$ is integer. So for $n=k$ it is true, so now comes time to proof it for $k+1$, $(2(n+1))!=(2n+2)!$, which is equal to $$1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2),$$ second term would be $$2^{n+1}=2 \times 2^n$$ Finally if we divide $(1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2)$ by $2^{n+1}=2 \times 2^n$,and consider that,$(2n)!/(2^n)$ is integer, we get $(2n+1) \times (2n+2)/2=(2n+1) \times 2 \times (n+1)/2$, we can cancel out $2$, we get $(2n+1)(n+1)$ which is definitely integer. I am curious this so simple? Does it means that I have proved correctly?

Answer 1

Yes, you have proved it correctly. Indeed the proof is not difficult. If you need to do a formal induction, fine. But the result becomes obvious if you just expand, say, $(2\cdot 5)!$. It is clear that you pick up at least five $2$'s.

If you do need to write out a formal induction, it could be written out somewhat more clearly. For example, the phrase "so for $n=k$ it is true" is not clear. I assume you mean that "so if for $n=k$ it is true." We now write out a proof.

The result is obviously true for $n=1$. We show that if it is true for $n=k$, it is true for $n=k+1$.

Note that $$(2\cdot (k+1))!=(2k)!(2k+1)(2k+2).$$ By the induction assumption, $2^k$ divides $(2k)!$. It follows that $2^{k+1}$ divides $(2k)!(2)$, and therefore $2^{k+1}$ divides $(2k)!(2k+1)(2k+2)$.

Answer 2

Suppose there are $n$ distinct objects in a set $D$.

Consider a set $S$ containing $2$ copies of each element from $D$. Then $S$ has total $2n$ objects.

Total number of permuatations of these objects $=\dfrac{(2n)!}{(2!)^n}=\dfrac{(2n)!}{2^n}$.

Since number of permutations is an integer, therefore, $\dfrac{(2n)!}{2^n}$ is an integer.

Answer 3

From 1 to $2n$ there are exactly $n$ even numbers. Hence the product $1\cdots 2n=(2n)!$ is divisible by $2\cdot2\cdots 2 \hbox{ ($n$ times)}= 2^n$.