Show that $\binom{2n}{  n}$ is divisible by 2? 
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prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer 

Show that $\binom{2n}{ n}$ is divisible by 2?
Any help would be appreciated..
 A: This answer has been moved to this question, of which this question has been judged to be a duplicate.
A: There are several ways to show this. I gave this as a homework exercise once (after having given the theory for computing a binomial coefficient modulo two in terms of the binary expansions), and a student surprised me with
$$
{2n\choose n}={2n-1\choose n-1}+{2n-1\choose n}=2{2n-1\choose n-1}.
$$
IOW he used Pascal's triangle rule once and then the symmetry ${n\choose k}={n\choose n-k}$.
A: The number $\binom{2n}{n}$ counts the number of ways to choose a set of $n$ people from a set of $2n$ people.
For every set $S$ of $n$ people, there is a corresponding other set of $n$ people, the complement $S'$ of $S$. Note that the complement of the complement of $S$ is $S$, So the subsets of size $n$ can be divided into complementary pairs, and therefore the number of sets of size $n$ is even.
If this is too abstract, let our initial set be $\{1,2,3,4,5,6\}$. Thus $n=3$.  The set $\{1,2,3\}$ is paired with its complement, which is the set $\{4,5,6\}$. The set $\{2,3,5\}$ is paired with its complementary set $\{1,4,6\}$, and so on. This pairing divides the sets of size $3$ into "couples."  Thus the number of subsets is twice the number of couples, and in particular is even.
Remark: This sort of pairing argument can be used to prove, for example, that if the positive integer $n$ is not a perfect square, then the number of (positive) divisors of $n$ is even. We illustrate the idea with $n=24$. Pair two divisors $a$ and $b$ of $n$ if $ab=n$. So $1$ is paired with $24$, $2$ is paired with $12$, $3$ is paired with $8$, $4$ is paired with $6$. Now the divisors of $24$ have been divided into couples, so $24$ has an even number of divisors. The same idea works for any non-square $n$.
The method breaks down when $n$ is a perfect square, like $36$. For then there is nobody to pair poor $6$ with. The other divisors are happily (?) off in couples, so the total number of divisors of $36$ is odd.  The same idea can be used to show that the total number of divisors of any perfect square is odd.
A: An algebraic approach:
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
