Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I show the following?

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

share|cite|improve this question
1  
Use the definition of factorial. – Chris Eagle Oct 10 '12 at 8:02
    
It s obvious, since $\frac{n-k}{(n-k)!}=\frac{1}{(n-(k+1))!}$ – S. Snape Oct 10 '12 at 8:04
    
@BabakSorouh, why? – idealistikz Oct 10 '12 at 8:11
    
@idealistikz: The Brian's is the complete answer. – S. Snape Oct 10 '12 at 8:19
up vote 3 down vote accepted

It’s just algebra:

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

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

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.