Proof of combination equation

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March 10th 2013, 08:21 PM
Paulo1913

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
March 11th 2013, 04:42 AM
veileen

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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