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Is there any way to simplify this?

$$\sum_{k=1}^{b}\left (\frac{n!}{k!(n-k)!}\right)$$

Edit: Assume that $b \le n$

(Side note: relates to my previous question.)

Community
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Geoffrey De Smet
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    Unless $b=n,n/2$ or other nice values, there is no way to simplify this sum. As in, there is no closed form expression for it. – Alex R. May 07 '14 at 15:04
  • I wonder if you might have intended $\displaystyle\sum_{k=0}^b$ rather than $\displaystyle\sum_{k=1}^b$? ${}\qquad{}$ – Michael Hardy May 07 '14 at 15:34
  • @MichaelHardy No, I don't want the empty collection included, but if it's included, I can easily count it out (+1) :) – Geoffrey De Smet May 07 '14 at 18:07

2 Answers2

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From \begin{align} \sum_{k=0}^{n} \binom{n}{k} = 2^{n} \end{align} then it is seen that (the case of $b=n$) \begin{align} \sum_{k=1}^{n} \binom{n}{k} = 2^{n} -1. \end{align} Now, if $b < n$ and integer, then \begin{align} \sum_{k=1}^{b} \binom{n}{k} = 2^{n} -1 - \sum_{k=b+1}^{n} \binom{n}{k}. \end{align} if $b > n$ and integer then \begin{align} \sum_{k=1}^{b} \binom{n}{k} = 2^{n} -1 + \sum_{k=n+1}^{b} \binom{n}{k} = 2^{n} -1 \end{align} since $\binom{n}{k}=0$ if $k >n$.

Leucippus
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This question has actually been asked quite a few times on this site: And the answer is that there is no closed form expression for the partial sum of binomial coefficients.

Lucian
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