Show that $\binom{2n}{  n}$ is divisible by 2? Possible Duplicate:
  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..
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 ${n\choose k}={n\choose n-k}$.
An algebraic approach:

$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$

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

$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$

$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$

$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$

$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$