1. ## Simplify factorials help!

Please can anyone help me to simplify
$(5n)!/(2n)!(3n)!$

Thanks, this is really bugging me.

2. Originally Posted by Frankler
Please can anyone help me to simplify
$(5n)!/(2n)!(3n)!$

Thanks, this is really bugging me.

You can expand it out and see what cancels.

$\frac{(5n)!}{(2n)!(3n)!} =$ $\frac{(5n)(5n-1)(5n-2)(5n-3)(5n-4)\cdot\cdot\cdot}{[(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdot\cdot\cdot][(3n)(3n-1)(3n-2)(3n-3)(3n-4)\cdot\cdot\cdot]}$

3. Originally Posted by Frankler
Please can anyone help me to simplify
$(5n)!/(2n)!(3n)!$

Thanks, this is really bugging me.
The (complete) gamma function $\Gamma(n)$ is defined to be an extension of the factorial to complex and real number arguments. It is related to the factorial by

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

So your expression can be stated as

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

4. 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 $
\frac{\Gamma(5 n+1)}{\Gamma(2n+1) \times \Gamma(3n+1)}
$

5. Originally Posted by Frankler
Please can anyone help me to simplify
$(5n)!/(2n)!(3n)!$

Thanks, this is really bugging me.
It's basically
choose 2n elements from 5n elements

$\binom{5n}{2n}=(5n)_{C_{(2n)}}=\frac{(5n)!}{(5n-2n)!(2n)!}$

6. Yes, that where I started with this problem, but does anyone know how it simplifies??

7. Originally Posted by Frankler
Yes, that where I started with this problem, but does anyone know how it simplifies??
Why are you so determined to try and simplify it?

8. Just to finish answering a question. Is there no way to simplify it? I'll just leave it then, thanks