How i can find closed form for the following sum?

Question

$\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{k}(n-k)!$

I converted to the below sum.Does anyone know of the closed form formula for this?

$\sum_{i=1}^{n}(-1)^{i+1}\frac{n!}{i!}$

Any hints or suggestions are welcomed.

Answer 1

That’s the number of derangements of $[n]$. There isn’t a nice closed form in the usual sense, but there is one if you allow the floor function:

$$\sum_{i=1}^n(-1)^{i+1}\frac{n!}{i!}=\left\lfloor\frac{n!}e+\frac12\right\rfloor\,.$$

Equivalently, it’s the integer nearest to $\frac{n!}e$.