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Math Help - Proof of combination equation

  1. #1
    Junior Member
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    Oct 2007
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    Proof of combination equation

    Hi,

    I am having trouble trying to show the following proof:

    C(2n,n)-C(2n,n-1)=(1/(n+1))*C(2n,n).

    C(2n,n) is equal to 2n!/(n!^2), and I have worked out C(2n,n-1) to be (2n)!/((n-1)!(n+1)! but I am not totally sure about that one, and if that is correct, I cannot seem to simplify it into the form required.

    Thanks for any help
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  2. #2
    Member
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    Mar 2011
    From
    Awetuouncsygg
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    Re: Proof of combination equation

    \textup{C}_{2n}^n-\textup{C}_{2n}^{n-1}=\textup{C}_{2n}^n-\frac{(2n)!}{(n-1)!\cdot (n+1)!}=

    =\textup{C}_{2n}^n-\frac{n\cdot (2n)!}{n\cdot (n-1)!\cdot  n! \cdot (n+1)}=\textup{C}_{2n}^n-\frac{n\cdot (2n)!}{n!\cdot  n! \cdot (n+1)}=

    =\textup{C}_{2n}^n-\frac{n}{n+1}\cdot \frac{ (2n)!}{n!\cdot  n!}=\textup{C}_{2n}^n-\frac{n}{n+1}\textup{C}_{2n}^n=

    =\left (1- \frac{n}{n+1} \right )\textup{C}_{2n}^n=\frac{n+1-n}{n+1}\textup{C}_{2n}^n=\frac{1}{n+1}\textup{C}_{  2n}^n
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