Number of permutations with 2 fixed points and 3 cycles Number of permutations with 2 fixed points and 3 cycles of {1,2,...,n}.
I ran into a problem with this. I know how to do it for 2 cycles, but for 3 it gets very complicated.
What I got: $$ \sum_{k=2}^{n/3} i^2 {n \choose k}*(k-1)!*\sum_{k=2}^{(n-k)/2} {n - k \choose l}*(l-1)!*(n-k-l-1)! $$
But I am kinda lost. Any help is appreciated.
 A: The species of  permutations with fixed points and  cycle count marked
is
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{V}\mathfrak{C}_{=2}(\mathcal{Z})
+ \mathcal{V}\mathfrak{C}_{=3}(\mathcal{Z})
+ \mathcal{V}\mathfrak{C}_{=4}(\mathcal{Z})
+ \cdots).$$
This gives the generating function
$$G(z, u, v) = 
\exp\left(uz + v\frac{z^2}{2} +
v\frac{z^3}{3} +
v\frac{z^4}{4} +
v\frac{z^5}{5} + \cdots\right)$$
which is
$$G(z, u, v) =
\exp\left(uz-vz + v\log\frac{1}{1-z}\right)
= \exp(uz-vz) \frac{1}{(1-z)^v}.$$
Now we have two fixed points so we get
$$[u^2] G(z, u, v)
= [u^2] \exp(uz) \exp(-vz) \frac{1}{(1-z)^v}
\\ = \frac{z^2}{2} \exp(-vz) \frac{1}{(1-z)^v}.$$
When  extracting the  count  for  three cycles  the  first version  is
actually more useful here and we get
$$[u^2] [v^3] G(z, u, v) =
[v^3] \frac{z^2}{2}
\exp\left(-vz + v\log\frac{1}{1-z}\right)$$
This finally yields
$$H(z) = \frac{z^2}{12}
\left(-z + \log\frac{1}{1-z}\right)^3.$$
We have by inspection that this is the species
$$\mathfrak{P}_{=2}(\mathfrak{C}_{=1}(\mathcal{Z}))
\mathfrak{P}_{=3}(\mathfrak{C}_{\ge 2}(\mathcal{Z})).$$
This generating function works when $n\ge 2+3\times 2=8$ since this is
the minimum number  of elements that we need for  two fixed points and
three cycles.
Starting at $n=8$ we get  the sequence of coefficients from the EGF
$H(z):$
$$ 420, 7560, 107100, 1453760, 20041560, 286949520, \ldots$$
The initial segment can be  verified with the following admittedly not
optimized Maple code:

with(combinat);

pet_disjcyc :=
proc(p)
local dc, pos;

    dc := convert(p, 'disjcyc');

    for pos to nops(p) do
        if p[pos] = pos then
            dc := [op(dc), [pos]];
        fi;
    od;

    dc;
end;


Q :=
proc(n)
    option remember;
    local perm, count, disjc;

    count := 0;

    perm := firstperm(n);
    while type(perm, `list`) do
        disjc := map(p->a[nops(p)],
                     pet_disjcyc(perm));

        if nops(disjc) = 5 then
            if degree(mul(q, q in disjc), a[1]) = 2 then
                count := count + 1;
            fi;
        fi;


        perm := nextperm(perm);
    od;

    count;
end;

Extracting coefficients we obtain 
$$n! [z^n] H(z)
= \frac{1}{12} n! [z^{n-2}]
\left(-z + \log\frac{1}{1-z}\right)^3
\\ = \frac{1}{12} n! [z^{n-2}]
\left((-z)^3 + 3(-z)^2 \left(\log\frac{1}{1-z}\right)
\\ + 3(-z) \left(\log\frac{1}{1-z}\right)^2
+ \left(\log\frac{1}{1-z}\right)^3\right).$$
This gives for $n\ge 8$ the closed form
$$\frac{1}{12} n!
\left( 3 \frac{1!}{(n-4)!} \left[n-4\atop 1\right]
- 3 \frac{2!}{(n-3)!} \left[n-3\atop 2\right]
+ \frac{3!}{(n-2)!} \left[n-2\atop 3\right]\right).$$
This finally yields
$$6 {n\choose 4} \left[n-4\atop 1\right]
- 3 {n\choose 3} \left[n-3\atop 2\right]
+ {n\choose 2} \left[n-2\atop 3\right].$$
This immediately reveals itself to be inclusion-exclusion. Reading
from the right we choose two  fixed points and combine them with three
cycles. Now the permutations with three fixed points have been counted
${3\choose  2}$  times  so  we  subtract  that  amount.   Finally  the
permutations  with  four fixed  points  have  been counted  ${4\choose
2}-3{4\choose 3}  = 6 - 12  = -6$ times so  we add that back  in for a
contribution of zero,  leaving a zero factor on  the permutations with
three and four fixed points. We  do not have to worry about five fixed
points because with  these we cannot represent a  permutation on $n\ge
8$  elements (in  fact this  is the  meaning of  the term  $(-z)^3$ in
$H(z)$ which contributes  a correction factor of $-10$  when $n=5$ and
indeed ${5\choose 2}- 3{5\choose 3} + 6{5\choose 4} = 10$).
