Simplifying the Factorial: How is it Done?

[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

<br /> (2n)! = (2n)(2n-1) \cdots (n+1) n!<br />

so things simply cancel.

 

[nitai108]

The original denominator is

<br /> (2n)! = (2n)(2n-1) \cdots (n+1) n!<br />

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.

<br /> \begin{align*}<br /> \frac{n!}{(2n)!} &amp; =\frac{4!}{8!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot \boxed{5} \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\<br /> &amp; = \frac{1}{8 \cdot 7 \cdot 6 \cdot \boxed{5}}= \frac{1}{(2n)\cdots (n+1)}<br /> \end{align*}<br />

Basically, when you write out the factorials in numerator and denominator, the final n factors cancel. Hope this helps.