De-arrangement in permutation and combination This article talks about de-arrangement in permutation combination.

Funda 1: De-arrangement
If $n$ distinct items are arranged in a row, then the number of ways they can be rearranged such that none of them occupies its original position is,
  $$n! \left(\frac{1}{0!} – \frac{1}{1!} + \frac{1}{2!} – \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!}\right).$$
Note: De-arrangement of 1 object is not possible.
$\mathrm{Dearr}(2) = 1$; $\mathrm{Dearr}(3) = 2$; $\mathrm{Dearr}(4) =12 – 4 + 1 = 9$; $\mathrm{Dearr}(5) = 60 – 20 + 5 – 1 = 44$.

I am not able to understand the logic behind the equation. I searched in the internet, but could not find any links to this particular topic.
Can anyone explain the logic behind this equation or point me to some link that does it ?

A while back, I posted three ways to derive the formula for derangements. Perhaps reading those might provide some insight into the equation above.
