Number of surjective functions where $f(a) = b$

Question

Let $A,B$ be groups such that $\left | A \right | = 7$ $\left | B \right | = 5$ and let $a\in A$ and $b\in B$

Find the number of onto functions $f:A \mapsto B$ where $f(a) = b$

Using the inclusion - exclusion principle I found that all the surjective functions from $A$ to $B$ is $16,800$

My question is, should I treat somehow that $f(a) = b$ for $a\in A$ and $b\in B$?

Does it change the way I treat the question somehow? thanks.

Answer 1

You've over-counted, since you've also counted those surjections $f : A \to B$ such that $f(a) \ne b$.

To count it properly, note that there are two possible cases. Either:

• $f(x) \ne b$ for any $x \ne a$. In this case, $f$ is uniquely determined by its restriction to a surjection $A \setminus \{ a \} \to B \setminus \{ b \}$; or
• $f(x) = b$ for some $x \ne a$. In this case, $f$ is uniquely determined by its restriction to a surjection $A \setminus \{ a \} \to B$.

So you need to count the number of surjections $A \setminus \{ a \} \to B \setminus \{ b \}$ and the number of surjections $A \setminus \{ a \} \to B$. The sum of these two values is what you seek.

Answer 2

Hint: Of all the surjective functions from $A$ onto $B$, what fraction have $f(a)=b$?