The number of permutations with specified number of cycles and fixed points

I'm interested in the number of permutations for a specified number of fixed points and cycles.

Suppose we are in $S_n$. For any permutation in $S_n$, let $h$ be the number of changed points (the complement of fixed points) and $N$ be the number of cycles of the permutation with length no less than 2 (number of cycles for the changed points). Therefore, for a given $h$, the number of cycles $N$ can range from $1$ to $\lfloor h/2\rfloor$.

So my question is how many permutations are there in $S_n$ for a given $h$ and $N$? If the exact expression is complicated, can we derive a concise and sharp upper bound of the number in terms of $n,h,N$?

A permutation can be viewed as a set of fixed points together with a derangement. The exponential generating function for sets of fixed points is $e^x$. If we weight each fixed point by $f$ then the generating function for weighted sets of fixed points is $e^{fx}$. The exponential generating function for derangements is $D(x)=e^{-x}/(1-x)$. Since each derangement is a set of cycles of length at least 2, if we weight each cycle of length at least 2 by $c$ then the exponential generating function for weighted derangements is $D(x)^c$. Thus the number of permutations in $S_n$ with $i$ fixed points (and therefore $n-i$ changed points) and $N$ cycles of length at least 2 is the coefficient of $\displaystyle f^ic^N\frac{x^n}{n!}$ in $$e^{fx}D(x)^c = \frac{e^{(f-c)x}}{(1-x)^c}.$$ From this generating function you can easily get an explicit formula.
The answer is $$\binom nh d_{h,N},$$ where $d_{h,N}$ is the number of derangements of size $h$ with $N$ cycles. By inclusion-exclusion we have: $$d_{h,N} = \sum_{i=0}^N (-1)^i\binom hi c(h-i,N-i),$$ where $c(\cdot,\cdot)$ are (unsigned) Stirling numbers of 1st kind.
P.S. Numbers $d_{h,N}$ are called associated Stirling numbers of first kind and listed in the sequence A008306 in the OEIS.
• Thank you! Is there any good upper bound for $d_{h,N}$? – neverevernever Sep 10 '18 at 17:09