Proving $\frac{((2n)!)^2(i)!(j)!}{((n)!)^2(2i)!(2j)!}$ is an integer I have verified this for many values of $n$, but I have no idea how to prove it. Does anyone know how I could go about showing that:
$$\frac{((2n)!)^2(i)!(j)!}{((n)!)^2(2i)!(2j)!}$$
is an integer when $n$, $i$, and $j$ are all integers and $n \geq i,j$
 A: First note that
$$\frac{(2n)!^2i!j!}{n!^2(2i)!(2j)!}=\frac{n!\binom{2n}n}{i!\binom{2i}i}\cdot\frac{n!\binom{2n}n}{j!\binom{2j}j}\;,\tag{1}$$
so we need only show that each of the factors on the righthand side of $(1)$ is an integer.
For any $k$, $k!\binom{2k}k$ is the number of ways of selecting $k$ of the members of the set $\{1,2,\ldots,2k\}$ and arranging them in some order; let $S_k$ be the set of $k$-tuples that result from this process. 
Take any sequence $\langle r_1,\ldots,r_i\rangle\in S_i$. Let $\langle r_{i+1},\ldots,r_n\rangle$ be any $(n-i)$-tuple of elements of the set $\{1,\ldots,2n\}\setminus\{r_1,\ldots,r_i\}$. There are $(n-i)!\binom{2n-i}{n-i}$ such $(n-i)$-tuples, and there are $\binom{n}i$ ways to merge one of them with $\langle r_1,\ldots,r_i\rangle$, so each member of $S_i$ is a substring of $(n-i)!\binom{2n-i}{n-i}\binom{n}i$ members of $S_n$. Thus,
$$\frac{n!\binom{2n}n}{i!\binom{2i}i}=\frac{|S_n|}{|S_i|}=(n-i)!\binom{2n-i}{n-i}\binom{n}i$$
is an integer, and similarly
$$\frac{n!\binom{2n}n}{j!\binom{2j}j}=\frac{|S_n|}{|S_j|}=(n-j)!\binom{2n-j}{n-j}\binom{n}j\;.$$
A: First lets define $f(n)$ to be equal to the product of the first n positive odd integers. We have $f(n)=\frac{(2n)!}{2^n (n)!}$ (explanation here). Logically, if $n \geq a$, then $\frac{f(n)}{f(a)}$ is an integer because all of the terms of $f(a)$ exist in $f(n)$.
Since $f(n)=\frac{(2n)!}{2^n (n)!}$, we also have $\frac{(2n)!}{(n)!}=2^n f(n)$ and likewise $\frac{n!}{(2n)!}=\frac{1}{2^n f(n)}$
Now, we can rewrite our original expression as:
$$\frac{(2n)!}{(n)!}\frac{(2n)!}{(n)!}\frac{(i)!}{(2i)!}\frac{(j)!}{(2j)!} = 2^n f(n) \cdot 2^n f(n) \cdot \frac{1}{2^i f(i)} \cdot \frac{1}{2^j f(j)}$$
$$=\frac{2^n}{2^i} \frac{2^n}{2^j} \frac{f(n)}{f(i)} \frac{f(n)}{f(j)}$$
And since $n \geq i$ and $n \geq j$, each fraction is an integer and thus the product is an integer.
If you were interested in this, they maybe you would be interested in trying to help out with the problem that this was a subproblem of: Proving an identity involving factorials
A: let $V_{p}(a)$ is the exponent of the prime number $p$ in the prime factorization of $a$,then we know
$$v_{p}(n!)=[\dfrac{n}{p}]+[\dfrac{n}{p^2}]+[\dfrac{n}{p^3}]+\cdots=\sum_{i=1}^{\infty}[\dfrac{n}{p^k}]$$
so Now we only prove
$$2\sum_{k=1}^{\infty}[\dfrac{2n}{p^k}]+\sum_{k=1}^{\infty}[\dfrac{i}{p^k}]+\sum_{k=1}^{\infty}[\dfrac{j}{p^k}]\ge 2\sum_{k=1}^{\infty}[\dfrac{n}{p^k}]+\sum_{k=1}^{\infty}[\dfrac{2i}{p^k}]+\sum_{k=1}^{\infty}[\dfrac{2j}{p^k}]$$
let
$$x=\dfrac{n}{p^k},y=\dfrac{i}{p^k},z=\dfrac{j}{p^k}$$
then we only prove
$$2[2x]+[y]+[z]\ge 2[x]+[2y]+[2z],x\ge y,z$$
use $x=[x]+\{x\}$,then we only prove
$$2[2([x]+\{x\})]+[y]+[z]\ge 2[x]+[2[y]+2\{y\}]+[2[z]+2\{z\}]$$
$$\Longleftrightarrow 2[x]+2[2\{x\}]\ge [y]+[z]+[2\{y\}]+[2\{z\}]$$
It is clear,we onlt consider following two case
since $x\ge y,z$,then if 
$\{x\}>0.5$,and $[x]=[y]=[z]=m$,then
$$LHS=2m+2\cdot 1=2m+2,RHS=m+m+2=2m+2$$
if 
$\{x\}<0.5$,and $[x]=m,[y]=[z]=m-1,\{y\},\{z\}>0.5$,then we have
$$LHS=2m+0=2m,RHS=m-1+m-1+1+1=2m$$
