- #1

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

Thanks a lot.

- Thread starter nitai108
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- #1

- 14

- 0

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

Thanks a lot.

- #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.

[tex]

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

[/tex]

so things simply cancel.

- #3

- 14

- 0

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)!.The original denominator is

[tex]

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

[/tex]

so things simply cancel.

- #4

statdad

Homework Helper

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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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