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Factorial simplification

  1. Jun 16, 2009 #1
    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.
  2. jcsd
  3. Jun 16, 2009 #2


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    The original denominator is

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

    so things simply cancel.
  4. Jun 16, 2009 #3
    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)!.
  5. Jun 16, 2009 #4


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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.

    \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)}

    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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