TITLE: Does $\sum_{k=0}^{k=n} {n \choose k} k!$ have a closed form for integers $k,n$?
QUESTION [8 upvotes]: While doing research in computer system, I came across the following summation:
$$S_n = \sum_{k=0}^{n} {n \choose k} k! = \sum_{k=0}^{n} \frac{n!}{(n-k)!}$$
where both $n$ and $k$ are integers.
$S_n$ has an intuitive meaning: Suppose we have $n$ distinct numbers. Each time we choose $k$ numbers and permute them. What is the total number of different permutations?
Though I only need the value of $S_2 = 5$ and $S_3 = 16$, I am interesting in its general form.  I have tried myself and googled with keywords such as "counting permutations, Stirling numbers", however I still have no idea.
EDIT:
I am quite satisfied with the accepted answer which, as one of the first answers, claims that $S_n = \lfloor n! e \rfloor$ and presents some numerical results. However, if you want to know how to prove it, please refer to the answer given by @Omran Kouta and give it an upvote.

REPLY [11 votes]: In fact all the previous posts propose  something like $S_n=\lfloor n!e\rfloor$ but without an explicit proof. Here is  one.
Define $u_n=\sum_{k=0}^n\frac{1}{k!}$ and $v_n=u_n+\frac{1}{n!\cdot n}$ it is an easy task to prove that $(v_n)$ is decreasing. This implies that $u_{n+1}\leq e\leq v_n$ for every $n\geq 1$.
This gives the following inequality
$$\forall\,n\geq1\qquad S_n+\frac{1}{n+1}\leq n! e\leq S_n+\frac{1}{n}$$
or equivalently
$$
n!e-1\leq n!e-\frac{1}{n}\leq S_n<n!e
$$
But $S_n$ is an integer, so the previous inequality proves that $S_n=\lfloor n!e\rfloor$.