An identity on Rencontres numbers Let there be $n$ people with seats marked $1$ to $n$. Let $p_k$ be the number of arrangements such that exactly $k$ persons go to their designated seat (the $i$ th person is designated seat number $i$) and the remaining do not. 
Show that $$\sum_{k = 0}^n k*p_k = n!$$
 A: We have  for permutations with  fixed points marked  the combinatorial
class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\mathcal{Z}
+ \textsc{CYC}_{\ge 2}(\mathcal{Z})).$$
This gives the mixed generating function
$$G(z, u) = \exp\left(uz + \sum_{q\ge 2} \frac{z^q}{q}\right)
= \exp\left(uz-z + \log\frac{1}{1-z}\right)
\\= \frac{1}{1-z} \exp(uz-z).$$
The  desired  quantity  is  then  given  by  (the  term  $u^k  z^n/n!$
representing  a permutation  of $n$  elements with  $k$ fixed  points)
should contribute $k z^n/n!$)
$$n! [z^n]
\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}
\\ = n! [z^n] \left. \frac{1}{1-z} \exp(uz-z) z \right|_{u=1}
\\ = n! [z^n] \frac{z}{1-z}.$$
This is zero for $n=0$ and evaluates to
$$\bbox[5px,border:2px solid #00A000]{n!}$$
otherwise.
A: The sum counts the number of people who go their designated seat, summed over all permutations. There are $n!$ permutations and $n$ people, and a person is equally likely to go to any particular seat, so they go to their designated seat in a fraction $\frac1n$ of all permutations. Thus this sum is $n!\cdot n\cdot\frac1n=n!$.
A: Probabilistic solution.
Take a random permutation and let $X$ be a number of people that seat on their designate position. Let $X_i$ be $1$ if $i$-th wo/man is designate else $X_i=0$. By definition we have $$E(X) = \sum_{k=0}^n kP(X=k) = \sum_{k=0}^n k{p_k\over n!}$$ 
but clearly since $X = X_1+X_2+...+X_n$ we have $$E(X) = E(X_1)+...+E(X_n) = {1\over n}+...+{1\over n} = 1$$
and we are done.
