# Simplify sum with factorials

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.)

• 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 7 '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 7 '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 7 '14 at 18:07

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 $k >n$.
• b <= n, so it looks like I am still stuck with the non-closed form. – Geoffrey De Smet May 7 '14 at 15:17