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I have the problem to evaluate the following:

$ (2n)!\over 2^n(n!) $

Does this reduce to anything in particular?

I stuck it into a computer and it's

1: 1 2: 3 3: 15 4: 105 5: 945 6: 10395 

No pattern immediately apparent.

  • 0
    @LuigiPlinge OK.. My colleague's whiteboard has the same sequence written all over it, complete with lots of little hand-drawn trees, that's why I was asking $:)$ Also, for such things the [On-Line Encyclopedia of Integer Sequences](http://oeis.org/) is a great resource.2012-10-03

4 Answers 4

12

Since $(2n)! = (2n) \times (2n-1) \times \cdots \times 2 \times 1$. Split the product into products of even factors and odd factors: $ (2n)! = \prod_{m=1}^{n} (2m) \cdot \prod_{m=1}^{n} (2m-1) = 2^n \prod_{m=1}^n m \cdot \prod_{m=1}^{n} (2m-1) = 2^n n! \prod_{m=1}^{n} (2m-1) $ Therefore: $ \frac{(2n)!}{2^n n!} = \prod_{m=1}^n (2m-1) = (2n-1)!! $ where $m!!$ denotes double factorial.

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    Great, the book doesn't mention double factorials but this must be what they were alluding to. Thanks!2012-10-02
5

You won’t get a nice closed form, but there is another way to write it that is sometimes useful. Notice that

$\begin{align*}2^nn!&=\underbrace{2\cdot2\cdot2\cdot\ldots\cdot2}_n\cdot1\cdot2\cdot3\cdot\ldots\cdot n\\&=(2\cdot1)(2\cdot2)(2\cdot3)\dots(2\cdot n)\\&=2\cdot4\cdot6\cdot\ldots\cdot 2n\;,\end{align*}$

do some cancelling, and look at Qiaochu’s comment.

2

You can use the identity

$ (2n)! = \Gamma(2n+1) = {\frac {{2}^{2n} \Gamma \left( n + 1\right) \Gamma \left( n + \frac{1}{2} \right) }{\sqrt {\pi }}}\,.$

This leads to the simplification

$\frac{(2n)!}{2^n n!} = \frac{ 2^n \Gamma(n+\frac{1}{2})}{\sqrt{\pi}} = \frac{ 2^n (n-\frac{1}{2})!}{\sqrt{\pi}}\,.$

0

As Brian said, you will not get a nice closed form expression. You can also do a little algebra and arrive at an expression with gamma functions instead of factorials.

http://en.wikipedia.org/wiki/Gamma_function