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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!}$

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  • Please add the link to go to the page of Wikipeadia – Supriyo Apr 20 '14 at 04:21
  • Yes, the formula on the last line is where this comes from. –  Apr 20 '14 at 05:02
  • $\displaystyle \sum_{k=0}^{2n+1}\dfrac{(-1)^k}{k!}< e^{-1}=\sum_{k=0}^\infty \dfrac{(-1)^k}{k!}<\sum_{k=0}^{2n}\dfrac{(-1)^k}{k!}$

    Therefor, we have that identity.

     – hxthanh Apr 20 '14 at 06:38

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