Explaining cardinality of set of all functions domain and codomain I have the below question that is being asked of me. For both parts, it is my understanding that a set has a specific cardinality of A -> B if it is a bijection of A -> B.
From the question I think what (a) is asking is for the cardinality of all possible functions that map parts of A to B, which I don't think leverages the bijection. How would I go about identifying ALL of the functions withougt brute forcing every possible combination?
And for (b), since a bijection implies that it is one-to-one AND onto then would would the cardinality be 0?

 A: Your first paragraph makes no sense: $A\to B$ is not a notation that represents a cardinality. What is true is that sets $A$ and $B$ have the same cardinality if and only if there is a bijection from one to the other. And neither (a) nor (b) asks about the cardinalities of the sets $A=\{1,2\}$ and $B=\{a,b,c\}$.
For (a) you are supposed to calculate how many functions there are from $A$ to $B$. A function from $A$ to $B$ is a set of two ordered pairs of the form $\{\langle 1,x\rangle,\langle 2,y\rangle\}$, where $x$ and $y$ are elements of $B$, possibly the same. For example, $f=\{\langle 1,c\rangle,\langle 2,a\rangle\}$ and $g=\{\langle 1,b\rangle,\langle 2,b\rangle\}$ are functions with domain $A$ and codomain $B$ (i.e., functions from $A$ to $B$): in more familiar notation, $f$ is the function such that $f(1)=c$ and $f(2)=a$, and $g$ is the constant function $g(1)=g(2)=b$. Counting these functions is straightforward. When you build one of them, $\{\langle 1,x\rangle,\langle 2,y\rangle\}$, how many different choices are there for $x$? How many for $y$? In how many ways can you combine these choices to get different functions?
For (b) you are to count the one-to-one functions from $A$ to $B$. In other words, you need to exclude from your count the ones like $g$ above that are not one-to-one. Now when you construct one, say $\{\langle 1,x\rangle,\langle 2,y\rangle\}$, you want $x$ and $y$ to be different members of $B$. Say you choose $x$ first and then $y$. How many choices are there for $x$? How many for $y$ once $x$ has been chosen? And how many different combinations can you get altogether?
Note that nowhere in these questions is there any mention of functions from $A$ onto $B$. There are in fact no such functions, but that fact has no bearing on the problems.
