Take the 2-minute tour ×
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.

I have the following sum:
$$\sum_{n=1}^k \frac{k!}{n!(k-n)!}, \quad k=9$$

wolfram alpha

It got simplified to $2^k-1$. How can I do it with math formulas? Thank you!

share|improve this question
3  
The function $\dfrac{k!}{n!(k-n)!}$ is often written $\binom k n$, and is called the binomial coefficient. –  Thomas Andrews Jul 11 '13 at 14:59
    
Thanks, it helped! –  Chelios Jul 11 '13 at 15:01
add comment

2 Answers

up vote 8 down vote accepted

$$ \sum_{n=0}^k a^n b^{k-n} \frac{k!}{n!(k-n)!} = (a+b)^k $$

share|improve this answer
add comment

$$\frac{k!}{n!(k-n)!}=\binom{k}{n},\binom{k}{0}=1$$ $$\sum_{n=0}^k\binom{k}{n}=2^k$$ $$\sum_{n=1}^k \frac{k!}{n!(k-n)!}=-1+\binom{k}{0}+\sum_{n=1}^k \binom{k}{n}=-1+\sum_{n=0}^k\binom{k}{n}=-1+2^k$$

share|improve this answer
    
I believe that it’s your second line that OP was asking about. –  Lubin Jul 11 '13 at 17:05
add comment

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.