As the question states, does this equation hold true?
$\sum_{j=0}^n \sum_{E \in {n \choose j}} (-1)^{|E|}(n-|E|)! = \sum_{j=0}^n(-1)^j(n-j)!{n \choose j} $
From what I understand, this holds true at least until the $(n-j)!$ bit on the right side of the equation, but I don't understand where the ${n \choose j}$ comes from. Can someone enlighten me?
For clarity: This is part of a demonstration to prove that the proportion of derangements of n over the number of permutations of n is $\frac1e$
On the left-hand side, the inner sum should probably be over $E \in \binom{[n]}{j} $, which would be over all subsets of $[n]=\{1,2,...,n\}$ of size $j$.
The right-hand side comes from observing that the summand only depends on $|E|$, and not the actual elements of $E$. Hence for each fixed $j$, the summands are identical. As there are $\binom{n}{j}$ choices for the set $E$, we pick up this additional factor.