Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
Have you noticed that each term divides the next term? –  Qiaochu Yuan Oct 2 '12 at 2:32
    
Thanks to everyone for your comments and answers. –  Luigi Plinge Oct 2 '12 at 3:02
    
Were you computing the possible arrangements of a commutative, non-associative, operator over n terms, by any chance? –  Philippe Oct 2 '12 at 9:34
    
@Philippe No, it's just an exercise from a book. –  Luigi Plinge Oct 2 '12 at 22:09
    
@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 is a great resource. –  Philippe Oct 3 '12 at 8:02

4 Answers 4

up vote 11 down vote accepted

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.

share|improve this answer
    
Great, the book doesn't mention double factorials but this must be what they were alluding to. Thanks! –  Luigi Plinge Oct 2 '12 at 3:02

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.

share|improve this answer

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

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.