Please can anyone help me to simplify

$\displaystyle (5n)!/(2n)!(3n)!$

Thanks, this is really bugging me.

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- Apr 8th 2010, 12:17 PMFranklerSimplify factorials help!
Please can anyone help me to simplify

$\displaystyle (5n)!/(2n)!(3n)!$

Thanks, this is really bugging me. - Apr 8th 2010, 12:31 PMAnonymous1
I could be wrong about this, but I don't know how much simpler this can get...

You can expand it out and see what cancels.

$\displaystyle \frac{(5n)!}{(2n)!(3n)!} =$ $\displaystyle \frac{(5n)(5n-1)(5n-2)(5n-3)(5n-4)\cdot\cdot\cdot}{[(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdot\cdot\cdot][(3n)(3n-1)(3n-2)(3n-3)(3n-4)\cdot\cdot\cdot]}$ - Apr 8th 2010, 02:09 PMharish21
The (complete) gamma function $\displaystyle \Gamma(n)$ is defined to be an extension of the factorial to complex and real number arguments. It is related to the factorial by

$\displaystyle \Gamma(n) = (n-1)!$

So your expression can be stated as

$\displaystyle \frac{\Gamma(5 n+1)}{\Gamma(2n+1) \times \Gamma(3n+1)}$ - Apr 11th 2010, 03:08 AMFrankler
Thanks both of you. I've tried cancelling it, and got rid of the (3n)!, but I don't see how that helps...

And how do you simplify $\displaystyle

\frac{\Gamma(5 n+1)}{\Gamma(2n+1) \times \Gamma(3n+1)}

$ - Apr 11th 2010, 05:51 AMArchie Meade
- Apr 11th 2010, 10:08 AMFrankler
Yes, that where I started with this problem, but does anyone know how it simplifies??

- Apr 12th 2010, 01:57 AMmr fantastic
- Apr 13th 2010, 02:58 AMFrankler
Just to finish answering a question. Is there no way to simplify it? I'll just leave it then, thanks :)