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

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

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.

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

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