discrete maths/ cardinality of sets/ functions Let
$n > m \geq 1$, $M=\{1,...,m\}$ and $N=\{1,...,n\}$. Furthermore, define the sets of all bijective mappings $f: N \to N$ with $F_n$.
Define
$A:=\{(a_1, \dots, a_n) | \{a_1,...,a_n\} \subset N\}$,
$B := \{(a_1, \dots, a_n) | \{a_1,...,a_n\} = N\}$,
$C := \{\{a_1, \dots, a_n\} | (a_1,...,a_n) \in A \text{/} B \}$

State formulas for $|A|, |B|, |C|$ and order them by size.
How many $f \in F_n$ are there such that $f(M)=M$ (meaning that $f(i) \leq m$ is fulfilled for $i \leq m$)?
Honestly I'm a little bit lost here, I know that I have to apply the binomial coefficient, but I do not know how to start. Don't quite understand the use of parentheses in the sets A/B/C. What exactly is $(a_1, \dots, a_n)$? Does $\{a_1,...,a_n\}$ just denote the elements (or a subset thereof) of N?
 A: Collating comments:
The parentheses notation is typically used for sequences or ordered-tuples of numbers, which are formally defined as functions from an index set of the form $\{1,2,\dots,k\}$ to whatever codomain set.  One can thus refer to sequences and functions interchangeably depending on preference.  I will refer to these as functions for the remainder of the post.
$A$ is talking about the set of functions $N\to N$ and is of size $n^n$.  Side note: Some authors will notate the set of functions $A\to B$ as $B^A$ which has the nice benefit of being able to say $|B^A|=|B|^{|A|}$
$B$ is talking about the set of functions $N\to N$ such that the range is the entirety of the codomain, i.e. here since $N$ is finite we have these are precisely the bijections $N\to N$.  Earlier in your post you were given the notation $F_n$ for this.  Many other authors prefer the notation $S_n$ in reference to the symmetric group.  There are $n!$ such functions.
$C$ if written as is is either talking about the number of possible ranges of non-bijective functions $N\to N$ which would be all possible subsets of $N$ apart from the empty set and apart from the set $N$ itself.  In the case of $n\geq 2$ that would be $2^n-2$ (and in the case of $n=0,1$ would be zero).
If $C$ were a typo and meant to be $\{(a_1,\dots,a_n)|(a_1,\dots,a_n)\in A\setminus B\}$ instead, that is the set of functions $N\to N$ who are not bijections, that would be the difference of the previous two, $n^n-n!$.
Typically there will be many more non-bijections than bijections, making the hierarchy here $|B|\leq |C|\leq |A|$, though some small values of $n$ may mess with that.

As for counting those $f\in F_n$ satisfying $f(M)=M$, those would be the bijective functions such that the first $m$ positions are occupied by a permutation of the elements of $M$ and the remaining positions are occupied by a permutation of what remains.  It will indeed be $m!(n-m)!$ such functions.
