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.)

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 $k >n$.
