Factorial simplification

[nitai108]

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

\frac{n!}{(2n)!} = \frac{1}{(2n)(2n-1)...(n+1)}


Thanks a lot.

[statdad]

The original denominator is


(2n)! = (2n)(2n-1) \cdots (n+1) n!


so things simply cancel.

[nitai108]

The original denominator is


(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]

Think about the meaning of (2n)! . It contains all the integers from 2n down to 1 . When you write out the entire factorial you must write down each one of those integers, and n + 1 is one of them.

As a specific (but small enough to write down) example, look what happens for n = 4 . This clearly means n+1 = 5 , which is the number I've placed in a box.


\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 n factors cancel. Hope this helps.