# binomials

• October 2nd 2008, 12:24 PM
oxrigby
binomials
I have been set this question and don't know wherto begin to get powers from..

Show that,, $\frac{(2n)!}{n!}=2^n.1.3.5...(2n-1)$
• October 2nd 2008, 12:56 PM
Plato
$\left[ {2n} \right]! = \left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)\left( {2n - 3} \right)\left( {2n - 4} \right) \cdots (4)(3)(2)(1)$
$\left[ {2n} \right]! = 2\left( n \right)\left( {2n - 1} \right)2\left( {n - 1} \right)\left( {2n - 3} \right)2\left( {n - 2} \right) \cdots 2(2)(3)2(1)(1)$, factor out 2's.
$\left[ {2n} \right]! = 2^n \left( n \right)\left( {2n - 1} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {n - 2} \right) \cdots (2)(3)(1)(1)$, there are n 2's. Do you see a factor of n!?
$\left[ {2n} \right]! = 2^n \left( {2n - 1} \right)\left( {2n - 3} \right)\left( {n - 2} \right) \cdots (3)(1)(n!)$
• October 2nd 2008, 01:09 PM
oxrigby
where did the n from your 3rd line come from?
• October 2nd 2008, 01:14 PM
Plato
Quote:

Originally Posted by oxrigby
where did the n from your 3rd line come from?

2 times itself n times is $2^n$.
• October 2nd 2008, 01:25 PM
oxrigby
got it