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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!

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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
up vote 8 down vote accepted

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

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

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I believe that it’s your second line that OP was asking about. – Lubin Jul 11 '13 at 17:05

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