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Binomial[2 n, 2 a]/Binomial[n, a]^2

I come across this expression, but no simplification command is working.
Can it be simplified further?

george2079
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Shamina
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1 Answers1

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After expanding into factorials, there's no generic opportunity to cancel anything. It's a choice between several equivalent forms, like

Binomial[2 n,n] / (Binomial[2 a, a] Binomial[2(n-a), n-a])

which may, depending on your problem, provide some further insights. But I think your form or mine hit the limit of what one would consider "simplifying" unless $n$ and/or $a$ are known.

Note that $$(2n)! = n! 2^n (2n-1)!!,$$ so one can cancel out one pair of factorials in $$\binom{2n}n = 2^n\frac{(2n-1)!!}{n!},$$ if they really want. Normally this would not be called a simplification but I've seen many problems where it lead to one. This could be applied up to three times in the other expression – again, depending on what you want to achieve.

The Vee
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  • Thanks a lot for your response. But still after the reference provided by george2079, in the comment is more close to simplification. Can you say something along those lines. It may of great help – Shamina Nov 18 '16 at 09:59
  • @Bonjour But what would you like to see in the output? You can make Mathematica expand the factorials for you as per the other link (no need to describe that anew), but that's just a matter of formatting. It's not going to be simpler than the ratio of the binomial coefficients you started with. Can you give an example of a formula with binomial coefficients and the kind of simplification you'd like to achieve? We can then try to reach that goal. – The Vee Nov 18 '16 at 21:37