Evaluate $\sum P(n,k)$

Question

It is known that $\sum_{k} {{n}\choose{k}} = 2^n$, and there is a combinatorial meaning assigned to it.

Is there some combinatorial meaning or simply algebraic manipulation to evaluate $\sum_{k} P(n,k)$, where $P(n,k)$ represents number of permutations.

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My work done so far: considering binomial inversion (failed), asignning to it a combinatorial meaning (failed), algebraic manipulation (failed), generating function (haven't tried).

Answer 1

Write:

$$S_n=\sum_{k=0}^{n}\frac{n!}{(n-k)!}=n!\sum_{k=0}^{\infty}\frac{1}{k!}-\sum_{k=n+1}^{\infty}\frac{n!}{k!}\tag{1}$$

Notice that the first summation on the right of $(1)$ is $n!e$. Also from direct term by term comparison with a geometric series of common ratio $1/(n+1)$ the subtraced summation in $(1)$ is, for $n\ge 1$:

$$\sum_{k=n+1}^{\infty}\frac{n!}{k!}\lt \sum_{k=n+1}^{\infty}\frac{1}{(n+1)^{k-n}}=\frac{1/(n+1)}{1-1/(n+1)}=\frac{1}{n}$$

so $S_n$ is always within $1/n$ of $n!e$ hence we may use the floor function:

$$S_n=\lfloor n!e\rfloor\tag{Answer}$$

This, of course, includes the empty permutation.

Answer 2

If $$P_{n,k}=\frac{n!}{(n-k)!}$$ then $$S_n=\sum_{k=1}^n P_{n,k}=e\,\Gamma (n+1,1)-1$$ where appears the incomplete gamma function.

The $S_n$'s are in sequence $A007526$ in $OEIS$