Q:

Combinatorics Question - Permutations and Supersets

I had a question that seems pretty straightforward, but I can't seem to wrap my mind around it.
Let's say I have a bunch of elements in a set. {A, B, C, D, E}.
How many permutations are there of every subset of this set? For example, I want to come up with every permutation of these elements that uses all possible elements, as well as all possible subsets of all the elements.
My guess is that we have 5! permutations to use all elements, 4! permutations to use a set of 4, which there are 5 of, 3! permutations to use a set of 3, which there are (5 * 4) of, 2! permutations to use a set of 2, which there are (5 * 4 * 3) of, and 1! permutations to use a set of 1, which there are (5 * 4 * 3 * 2) of. 
Mathematically, how can this expression be generalized to a set of size n?
Thanks!

A:

The number of permutations of every subset of a set of size $n$ is $\lfloor n!e\rfloor$. This is because
$$
\sum_{k=0}^n\underbrace{\binom{n}{k}}_{\substack{\text{number of}\\\text{subsets of}\\\text{size $k$}}}\underbrace{k!\vphantom{\binom{n}{k}}}_{\substack{\text{number of}\\\text{permutations}\\\text{on a subset}\\\text{of size $k$}}}
=\sum_{k=0}^n\frac{n!}{(n-k)!}
=\sum_{k=0}^n\frac{n!}{k!}
=\lfloor n!e\rfloor
$$

