How many injective functions $f: A \to B$ satisfy $f(a_1) = b_1$ or $f(a_2)=b_2$?

Question

I've been stuck on this question for hours, and am having trouble trying to start this question. If anyone could help, that would much appreciated.

The question is Let $A = \{a_1, a_2, a_3, a_4\}$ and $B = \{b_1, b_2, b_3, b_4, b_5\}$ (where $|A| =4$ and $|B|=5$), how many injective functions $f: A \to B$ satisfy $f(a_1) = b_1$ or $f(a_2)=b_2$?

Answer 1

Define $F_1 = \{f\colon A\to B | f(a_1) = a_1, f \text{ injective} \}$. Define $F_2$ similarly for $f(a_2) = b_2$. Then use the fact that functions in $F_1$ are injective to find the cardinality of $F_1$ and similarly for $F_2$. Once you've done this, find the cardinality of $F_1 \cap F_2$. Then the total number of injective functions $f\colon A\to B$ satisfying $f(a_1) = b_1$ or $f(a_2) = b_2$ is $|F_1| + |F_2| - |F_1 \cap F_2|$ by the inclusion exclusion principle.

Answer 2

Case 1: $a_1$ is mapped to $b_1$

There are four ways to map $a_2$; to $b_2, b_3, b_4,$ of $b_5$.

There are three ways to map $a_3$. (To any of the three remaining values).

There are two ways to map $a_4$. (To any of the two remaining.)

That's 24 ways.

Case 2: $a_1$ is not mapped to $b_1$

Then $a_2$ must map to $b_2$.

There are three ways to map $a_1$; to $b_3, b_4, b_5$

After $a_1$ and $a_2$ are mapped; there are three remaining ways to map $a_3$.

After $a_1, a_2,$ and $a_3$ are mapped there are two remaining ways to map $a_4$.

That's $18$ ways.

That's $24 + 18 = 42$ ways. I'm going to need a lot more pinkies.