For the first part of the question with $Q(n, k, m)$ the number of
cycles of size $m$ among permutations of $[n]$ having $k$ cycles we
get the species
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=k}(
\textsc{CYC}_{\lt m}(\mathcal{Z})
+ \mathcal{U}\times \textsc{CYC}_{= m}(\mathcal{Z})
+ \textsc{CYC}_{\gt m}(\mathcal{Z})).$$
This yields the bivariate generating function
$$G(z, u) = \frac{1}{k!}\left(\log\frac{1}{1-z}-\frac{z^m}{m}
+ u\frac{z^m}{m}\right)^k.$$
The desired statistic has generating function
$$\left.\frac{\partial}{\partial u} G(z, u)\right|_{u=1}
\\ = \left. \frac{1}{(k-1)!}\left(\log\frac{1}{1-z}-\frac{z^m}{m}
+ u\frac{z^m}{m}\right)^{k-1} \frac{z^m}{m} \right|_{u=1}
\\ = \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1} \frac{z^m}{m}.$$
Extracting coefficients we obtain
$$n! [z^n] \frac{z^m}{m} \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \frac{n!}{m} [z^{n-m}] \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \frac{n!}{m (n-m)!} (n-m)! [z^{n-m}] \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}.$$
Divide by ${n\brack k}$ for the expectation
$$\bbox[5px,border:2px solid #00A000]{
\frac{n!}{m (n-m)!} {n-m\brack k-1} {n\brack k}^{-1}.}$$
For the proof that these sum to $k$ which must hold by first
principles we observe that we need $n-m\ge k-1$ or $n+1-k\ge m$ and
obtain the claim
$${n\brack k}^{-1}
\sum_{m=1}^{n+1-k} \frac{n!}{m (n-m)!} {n-m\brack k-1} = k.$$
The EGF of the sum is
$$\sum_{n\ge k} \frac{w^n}{n!}
\sum_{m=1}^{n+1-k} \frac{n!}{m (n-m)!}
(n-m)! [z^{n-m}] \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \sum_{n\ge k} w^n
\sum_{m=1}^{n+1-k} \frac{1}{m}
[z^{n}] z^m \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}.$$
Now we may actually extend the inner sum to infinity because when
$m\gt n+1-k$ we have $m+k-1\gt n$ and there is no contribution to
$[z^n].$ We get
$$\sum_{n\ge k} w^n
\sum_{m\ge 1} \frac{1}{m}
[z^{n}] z^m \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \sum_{n\ge k} w^n [z^n]
\sum_{m\ge 1} \frac{1}{m}
z^m \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \sum_{n\ge k} w^n [z^n]
\frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k}
\\ = \frac{1}{(k-1)!}
\left(\log\frac{1}{1-w} \right)^{k}.$$
Extracting coefficients we find
$${n\brack k}^{-1} n! [w^n] k \frac{1}{k!}
\left(\log\frac{1}{1-w} \right)^{k} =
{n\brack k}^{-1} k {n\brack k} = k$$
as claimed. For sanity check number two the claim is
$${n\brack k}^{-1}
\sum_{m=1}^{n+1-k} \frac{n!}{(n-m)!} {n-m\brack k-1} = n.$$
Re-using the computation from the first one yields
$$\sum_{n\ge k} w^n [z^n]
\sum_{m\ge 1}
z^m \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \sum_{n\ge k} w^n [z^n]
\frac{z}{1-z} \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z} \right)^{k-1}
\\ = \frac{w}{1-w} \frac{1}{(k-1)!}
\left(\log\frac{1}{1-w} \right)^{k-1}
= w \frac{d}{dw} \frac{1}{k!}
\left(\log\frac{1}{1-w} \right)^{k}.$$
Extracting coefficients we find
$${n\brack k}^{-1} n! [w^n] w \frac{d}{dw} \frac{1}{k!}
\left(\log\frac{1}{1-w} \right)^{k}
= {n\brack k}^{-1} n {n\brack k} = n$$
again as claimed. The first sanity check must hold because when we sum
the expectations of the number of cycles of all possible sizes we
should get $k$ cycles, which is constant in the problem. The second
must hold because if we sum the lengths times the number of cycles of
all possible sizes we should cover all of $n,$ also a constant here.
These formulae were verified using the cycle index $Z(S_n)$ of the
symmetric group with the following Maple script which is practicable
to about $n=45.$ For example with $n=20$ and $k=15$ (permutations of
twenty elements having fifteen cycles) we get for the total count of
cycles of lengths one to six the values
$$10995785640, 2640350580, 737990400, 191280600, 39070080, 4651200$$
and for the expectation we find
$$11.28998110, 2.710993931, 0.7577355487, 0.1963983683,
\\ 0.04011541140, 0.004775644215$$
and these do indeed sum to fifteen.
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;
Q :=
proc(n, k)
option remember;
local idx, conjclass, cyc, count;
count := table([seq(q=0, q=1..n+1-k)]);
if n=1 then
idx := [a[1]]
else
idx := pet_cycleind_symm(n);
fi;
for conjclass in idx do
if degree(conjclass) = k then
for cyc in indets(conjclass) do
count[op(1, cyc)] :=
count[op(1, cyc)]
+ degree(conjclass, cyc)
* lcoeff(conjclass);
od;
fi;
od;
[seq(n!*count[q], q=1..n+1-k)];
end;
QX :=
(n, k, m) -> n!/m/(n-m)! * abs(stirling1(n-m, k-1));
QXS := (n, k) -> [seq(QX(n, k, m), m=1..n+1-k)];
Remark. The above used the technique of annihilated coefficient
extractors (ACE), otherwise known as the substitution rule for formal
power series. There are several more examples at this MSE link
I and at this MSE
link II and also
here at this MSE link
III.
