This is a practice question and not HW
Q. Let $A=\{1,2,3,4,5,6,7\}$ and $B=\{v,w,x,y,z\}$. Determine the number of functions $f:A \longrightarrow B$ where
a) $f(A)=\{v,x\}$
Ans. $2!*S(7,2)$
I understand that $S(7,2)$ is the total number of onto function from $A$ to $B$ but why multiply by $2!$? I do see see that $f(x)=(v,w)$ has two items so it certainly has something to do with that but why the factorial and why multiply it. Some intuition would be greatly appreciated.
b) $|f(A)|=2$
here the answer is $(5C2)(2!)S(7,2)$ Again why the $2!$ and why $5C2$
$S(n,k)$, or ${n\brace k}$, to use my preferred notation, is the number of ways to divide $n$ labelled objects amongst $k$ unlabelled containers so that no container is empty. If we then attach distinct labels to the $k$ containers — the numbers $1$ through $k$, for instance — we can do it in $k!$ different ways, one for each permutation of the containers. Thus, there are $k!{n\brace k}$ ways to divide $n$ labelled objects amongst $k$ labelled containers in such a way that no container is empty. In the case of a function from an $n$-element set onto a $k$-element set, you’re assigning each of the $n$ individually identifiable elements of the domain to one of the $k$ individually identifiable elements of the range, so the number of functions is $k!{n\brace k}$, not $n\brace k$. In particular, in this problem it’s $2!{7\brace 2}$.
In part (b) you have $2!{7\brace 2}$ functions for each possible $2$-element subset of $B$. Since $|B|=5$, there are $\binom52$ such subsets, each contributing $2!{7\brace 2}$ functions, for a total of $\binom52\cdot2!\cdot{7\brace2}$ functions.