To start off, there is the well known formula that the number of surjections $S$ from a set $M=\{1, ... , m\}$ to $N=\{1, ... , n\}$ for all $m>n$ is:
$S = n^m - {n \choose 1}(n-1)^m + {n \choose 2}(n-2)^m - \dots \pm {n \choose n}(n-n)^m$
Now let us take a particular case where we have $N\subset M$. Given this condition, we can say that in some of these $S$ surjections, there are some elements that are being mapped to themselves, and hence one can also ask what are the number of "derangements" for which I will use the notation $D_n^m$.
As a more thorough definition let us take this:
The number of derangements $D_n^m$ from a set $M$ to the set $N\subset M$ is the number of surjections such that no element is mapped to itself
Now given that $D_n^m = S - A_n^m$ where $A_n^m$ is the number of surjections where at least one element is mapped to itself, I tried both the approaches of finding $D_n^m$ directly and of finding $A_n^m$ and subtracting it from $S$ but I did not seem to arrive at anything that made sense. I also tried to approach it from a Group Theory angle but I did not get far enough with that to make it worth mentioning.
Is there a way to modify the formula for $S$ to restrict it to either $D_n^m$ or $A_n^m$? Is the Combinatorial approach the best way to tackle this or would a Group Theory approach be better?