Am I on the right track? I am not sure about my reasoning…
Number of surjective functions from $A$ to $B$
$$A = \{1,2,3,4\} ; B = \{a,b,c\}$$
We must count the surjective functions, meaning the functions for which for all $b \in B$, $\exists~a \in A$ such that $f(a) = b$, $f$ being one of those functions. In order for a function $f:A\rightarrow B$ to be a surjective function, all 3 elements of $B$ must be mapped.
We need to count how many ways we can map those 3 elements. We will subtract the number of functions from $A$ to $B$ which only maps 1 or 2 elements of $B$ to the number of functions from $A$ to $B$ (computed in 4.c : 81).
Only 1 element of $B$ is mapped
The first $a \in A$ has three choices of $b \in B$. The others will then only have one. Total functions from $A$ to $B$ mapping to only one element of $B$ : 3.
Exactly 2 elements of $B$ are mapped
Similarly, there are $2^4$ functions from $A$ to $B$ mapping to 2 or less $b \in B$. However, these functions include the ones that map to only 1 element of $B$. So there are $2^4-3 = 13$ functions respecting the property we are looking for.
In the end, there are $(3^4) – 13 – 3 = 65$ surjective functions from $A$ to $B$.
Best Answer
$\left\lbrace{4\atop 3}\right\rbrace=6$ is the number of ways to partition $A$ into three nonempty unlabeled subsets. For each partition, there is an associated $3!$ number of surjections, (We associate each element of the partition with an element from $B$). Thus, the total number of surjections is $3! \times \left\lbrace{4\atop 3}\right\rbrace= 36.$