How many permutations with $k$ cycles and $j$ fixed points are there? Suppose I have the numbers $\{1,...,n\}$. I would like to know how many permutations have exactly $k$ cycles and $j$ fixed points. 
 A: Hint: Here the needed concept is $2-$associated Stirling numbers of the first kind. First choose your fix points and the remaining is that.
A: By  way   of  enrichment  here  is  an   alternate  formulation  using
combinatorial  classes. What  we present  below has  the  advantage of
permitting  the computation  of  a wide  variety  of statistics.   The
class of permutations with cycles and fixed points marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\times \mathcal{V} \times
\textsc{CYC}_{=1}(\mathcal{Z})
+ \mathcal{U}\times \textsc{CYC}_{=2}(\mathcal{Z})
\\ + \mathcal{U}\times \textsc{CYC}_{=3}(\mathcal{Z})
+ \mathcal{U}\times \textsc{CYC}_{=4}(\mathcal{Z})
+ \cdots).$$
This gives the generating function
$$G(z, u, v) = 
\exp\left(uvz + u\frac{z^2}{2} +
u\frac{z^3}{3} +
u\frac{z^4}{4} +
u\frac{z^5}{5} + \cdots\right)$$
which is
$$G(z, u, v) =
\exp\left(uvz-uz+u\log\frac{1}{1-z}\right).$$
Extracting first the coefficient on $[v^j]$ we obtain
$$\frac{u^j z^j}{j!} 
\exp\left(-uz+u\log\frac{1}{1-z}\right)
= \frac{u^j z^j}{j!} 
\exp(-uz) \exp\left(u\log\frac{1}{1-z}\right).$$
We extract the coefficient on $[u^k]$ in the next step:
$$[u^k] \frac{u^j z^j}{j!} 
\exp(-uz) \exp\left(u\log\frac{1}{1-z}\right)
\\ = \frac{z^j}{j!} [u^{k-j}]
\exp(-uz) \exp\left(u\log\frac{1}{1-z}\right)
\\ = \frac{z^j}{j!}
\sum_{q=0}^{k-j} \frac{(-1)^q z^q}{q!} 
\frac{1}{(k-j-q)!} \left(\log\frac{1}{1-z}\right)^{k-j-q}.$$
To conclude we extract the coefficient on $[z^n]$ (EGF) which yields
$$n! [z^n] \frac{z^j}{j!}
\sum_{q=0}^{k-j} \frac{(-1)^q z^q}{q!} 
\frac{1}{(k-j-q)!} \left(\log\frac{1}{1-z}\right)^{k-j-q}
\\ = n! \frac{1}{j!}
\sum_{q=0}^{k-j} \frac{(-1)^q}{q!} [z^{n-j-q}]
\frac{1}{(k-j-q)!} \left(\log\frac{1}{1-z}\right)^{k-j-q}
\\ = n! \frac{1}{j!}
\sum_{q=0}^{k-j} \frac{(-1)^q}{q! \times (n-j-q)!} 
\\ \times (n-j-q)! [z^{n-j-q}]
\frac{1}{(k-j-q)!} \left(\log\frac{1}{1-z}\right)^{k-j-q}
\\ = n! \frac{1}{j!}
\sum_{q=0}^{k-j} \frac{(-1)^q}{q! \times (n-j-q)!} 
\left[n-j-q\atop k-j-q\right].$$
This yields the closed form
$$\bbox[5px,border:2px solid #00A000]{
{n\choose j}
\sum_{q=0}^{k-j} {n-j\choose q} (-1)^q
\left[n-j-q\atop k-j-q\right].}$$
Note  that we  cannot  simply use  a  Stirling number  $\left[n-j\atop
k-j\right]$ as the second factor  because that would include potential
additional   fixed  points.   The   formula  accounts   for  this   by
inclusion-exclusion (PIE). 
The following Maple program shows  how to compute this statistic using
the cycle index $Z(S_n)$ of the symmetric group.

with(combinat);

pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;


ENUM :=
proc(n, j, k)
    option remember;
    local part, idx, res;

    res := 0; 

    if n=1 then
        idx := [a[1]];
    else
        idx := pet_cycleind_symm(n);
    fi;

    for part in idx do
        if degree(part) = k and
        degree(part, a[1]) = j then
            res := res + lcoeff(part);
        fi;
    od;

    n!*res;
end;

X := (n, j, k) -> binomial(n,j)
* add(binomial(n-j,q)*(-1)^q
      * abs(stirling1(n-j-q, k-j-q)), q=0..k-j);

