1
$\begingroup$

This question already has an answer here:

I saw on wikipedia that a formula for derangements is

$\text{Round}\left[\frac{n!}{e}\right]$

However, how did they arrive at this elegant formula?

Does it have to do with $ !n=n! \sum _{k=0}^n \frac{(-1)^k}{k!}$

$\endgroup$

marked as duplicate by user85798, colormegone, Claude Leibovici, WimC, Namaste Apr 20 '14 at 11:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Please add the link to go to the page of Wikipeadia $\endgroup$ – Dutta Apr 20 '14 at 4:21
  • $\begingroup$ Yes, the formula on the last line is where this comes from. $\endgroup$ – user98602 Apr 20 '14 at 5:02
  • $\begingroup$ $\displaystyle \sum_{k=0}^{2n+1}\dfrac{(-1)^k}{k!}< e^{-1}=\sum_{k=0}^\infty \dfrac{(-1)^k}{k!}<\sum_{k=0}^{2n}\dfrac{(-1)^k}{k!}$ Therefor, we have that identity. $\endgroup$ – hxthanh Apr 20 '14 at 6:38