We have a set $A$ consisting of $n$ elements.
Is there a closed form for the total number of subsets when you care about the order of the elements in the subsets?
Lets call the number of subsets $T(n)$
$$T(n) = \binom{n}{0}0!+ \binom{n}{1}1!+ \binom{n}{2}2!+ \cdots +\binom{n}{n}n! $$
Which is the same as
$$P(n,0) + P(n,1) + P(n,2) + \cdots + P(n,n)$$
The "related" sum of combinations have a nice closed form.
$$C(n,0) + C(n,1) + C(n,2) + \cdots + C(n,n) = 2^n$$
If we factor out a $n$ we get
$$T(n) = 1 + n\left[ \binom{n-1}{0}0!+ \binom{n-1}{1}1!+ \cdots +\binom{n-1}{n-1}(n-1)! \right]$$
and the recurrence equation
$$T(n) = 1 +nT(n-1)$$
Is there a closed form for the solution of this for particular n?
I know that in the limit
$$\lim_{n \to \infty}T(n) = e \cdot n!$$
Mathematica gives $e \times \Gamma(n+1,1)$ where $$\Gamma(a, z) = \int_z^{\infty} t^{a-1} e^{-t} dt$$ is the incomplete gamma function. It's not much of a simplification, I know, but I don't think $\Gamma(a, 1)$ has a nicer closed form.