Factorial simplification

**nitai108** · Jun16-09, 07:22 AM

Can somebody please explain to me this simplification and how it's done?

$LaTeX Code: \\frac{n!}{(2n)!}$ = $LaTeX Code: \\frac{1}{(2n)(2n-1)...(n+1)}$

Thanks a lot.

**statdad** · Jun16-09, 07:50 AM

The original denominator is

$LaTeX Code: (2n)! = (2n)(2n-1) \\cdots (n+1) n! $

so things simply cancel.

**nitai108** · Jun16-09, 08:54 AM

Originally Posted by statdad

The original denominator is

$LaTeX Code: (2n)! = (2n)(2n-1) \\cdots (n+1) n! $

so things simply cancel.

Thanks for the help. I still don't understand the (n + 1), where does it come from? I've tried to search the net, and my textbooks but I never found examples of (xn)!, only n! = n(n-1)!.

**statdad** · Jun16-09, 09:11 AM

Think about the meaning of

. It contains all the integers from

down to

. When you write out the entire factorial you must write down each one of those integers, and

is one of them.

As a specific (but small enough to write down) example, look what happens for

. This clearly means

, which is the number I've placed in a box.

$LaTeX Code: \\begin{align*} \\frac{n!}{(2n)!} & =\\frac{4!}{8!} = \\frac{4 \\cdot 3 \\cdot 2 \\cdot 1}{6 \\cdot \\boxed{5} \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}\\\\ & = \\frac{1}{8 \\cdot 7 \\cdot 6 \\cdot \\boxed{5}}= \\frac{1}{(2n)\\cdots (n+1)} \\end{align*} $

Basically, when you write out the factorials in numerator and denominator, the final

factors cancel. Hope this helps.