TITLE: Why is $\sum_{p \in S_n} 2^{c(p)}$ equal to $(n+1)!$?
QUESTION [15 upvotes]: It is obvious that $\sum_{p \in S_n} 1=n!$ because it is just counting how many permutations there are of $n$ symbols.
But I have also observed that $\sum_{p \in S_n} 2^{c(p)}=(n+1)!$, where $c(p)$ is the number of cycles of $p$. 
What is the combinatorial interpretation of this identity?
An example. In $S_3$ we have one permutation with 3 cycles, three permutations with 2 cycles and two with 1 cycle. Then $1\times 2^3+3\times 2^2+2\times 2^1=24=4!$

REPLY [8 votes]: The combinatorial class of permutations with the number of cycles marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(
\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z})
+ \mathcal{U}\times  \textsc{CYC}_{=2}(\mathcal{Z})
+ \mathcal{U}\times  \textsc{CYC}_{=3}(\mathcal{Z})
+ \cdots).$$
This gives the EGF
$$G(z, u) = \exp\left(uz+u\frac{z^2}{2} + u\frac{z^3}{3}+\cdots\right)
\\ = \exp\left(u\log\frac{1}{1-z}\right).$$
A permutation on $n$ elements and having $k$ cycles is represented by $u^k
\frac{z^n}{n!}.$ For the sum we set $u=2$ and obtain
$$H(z) = \exp\left(2\log\frac{1}{1-z}\right)
= \frac{1}{(1-z)^2}.$$
We then get for the answer
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^2}
= n! {n+1\choose 1} = (n+1)!.$$
This is the claim.
 Addendum. With two being replaced by $m$ we obtain
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^m}
= n! \times {n+m-1\choose m-1} = n! \times {n+m-1\choose n}.$$