Factorial simplification

  • Thread starter nitai108
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  • #1
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Can somebody please explain to me this simplification and how it's done?

[tex]\frac{n!}{(2n)!}[/tex] = [tex]\frac{1}{(2n)(2n-1)...(n+1)}[/tex]


Thanks a lot.
 

Answers and Replies

  • #2
statdad
Homework Helper
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The original denominator is

[tex]
(2n)! = (2n)(2n-1) \cdots (n+1) n!
[/tex]

so things simply cancel.
 
  • #3
14
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The original denominator is

[tex]
(2n)! = (2n)(2n-1) \cdots (n+1) n!
[/tex]

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)!.
 
  • #4
statdad
Homework Helper
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Think about the meaning of [tex] (2n)! [/tex]. It contains all the integers from [tex] 2n [/tex] down to [tex] 1 [/tex]. When you write out the entire factorial you must write down each one of those integers, and [tex] n + 1 [/tex] is one of them.

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

[tex]
\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*}
[/tex]

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

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