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How do I show the following?

$$\frac{n!}{(k+1)!(n-(k+1))!}=\frac{n-k}{k+1}\frac{n!}{k!(n-k)!} \text{ for } k=0,1,\ldots,n-1$$

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1  
Use the definition of factorial. –  Chris Eagle Oct 10 '12 at 8:02
    
It s obvious, since $\frac{n-k}{(n-k)!}=\frac{1}{(n-(k+1))!}$ –  Babak S. Oct 10 '12 at 8:04
    
@BabakSorouh, why? –  idealistikz Oct 10 '12 at 8:11
    
@idealistikz: The Brian's is the complete answer. –  Babak S. Oct 10 '12 at 8:19
    
Although you are not new, I wanted to remind you of a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also please give attribution to who wrote the problem (book, teacher, etc.) –  Noah Snyder Oct 10 '12 at 9:16

2 Answers 2

up vote 3 down vote accepted

It’s just algebra:

$$\begin{align*} \frac{n-k}{k+1}\cdot\frac{n!}{k!(n-k)!}&=\frac{(n-k)n!}{(k+1)k!(n-k)!}\\ &=\frac{(n-k)n!}{(k+1)!(n-k)!}\quad\text{since}(k+1)!=(k+1)k!\\ &=\frac{\color{red}{(n-k)}n!}{(k+1)!\color{red}{(n-k)}(n-k-1)!}\quad\text{since}(n-k)!=(n-k)(n-k-1)!\\ &=\frac{n!}{(k+1)!(n-k-1)!}\\ &=\frac{n!}{(k+1)!\big(n-(k+1)\big)!} \end{align*}$$

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@idealistikz: No; where do you think that there should be one? –  Brian M. Scott Oct 10 '12 at 8:23

$$\frac{n!}{(k+1)!(n-(k+1))!}=\frac{n!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)k!\frac{(n-k)!}{n-k}}=\frac{n-k}{k+1}\frac{n!}{k!(n-k)!}$$

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