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..

An algebraic approach:
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
