Simplifying Combination Formula: \binom{n}{\frac{n}{2}} using Factorial Formula

Aleoa
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I'm trying to simplify the combination defined as : \binom{n}{\frac{n}{2}}.

I did some calculations, starting from the factorial formula \frac{n!}{(\frac{n}{2})!(\frac{n}{2})!} and i found this form :

2^{n}(1-\frac{1}{n})(1-\frac{1}{n-2})(1-\frac{1}{n-4})...

but i don't know how to continue, can you help me ?
 
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Aleoa said:
but i don't know how to continue, can you help me ?
No, since you haven't said what your goal is. To me ##\binom{n}{\frac{n}{2}}## is already fine.
 
I want to characterize the behaviour of the formula as n become larger, so I'm trying to simplify it . I'm sorry for the template, next time i'll write it correctly. Thanks for the support
 
Aleoa said:
I want to characterize the behaviour of the formula as n become larger, so I'm trying to simplify it . I'm sorry for the template, next time i'll write it correctly. Thanks for the support
In this case I'd try where Stirling's approximation would get me.
 
Aleoa said:
I want to characterize the behaviour of the formula as n become larger, so I'm trying to simplify it . I'm sorry for the template, next time i'll write it correctly. Thanks for the support

As fresh_42 suggested, use Stirling's formula. Every student of probability should be thoroughly familiar with that formula, as it is used everywhere.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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