2012 AMC 10B Problem 24

Describe each song by the set of girls who like it. The set $\{A, B, J\}$ is forbidden, leaving $7$ possible labels: the empty set, three singletons, and the three pairs $AB$, $AJ$, $BJ$. The second condition says each of the three pair-labels is used by at least one of the four songs. Since there are four songs and three required labels, split by how the pair-labels are distributed.

Case 1: each pair-label used once, plus one other song. Choose which song is the extra one ($4$ ways), assign the three pair-labels to the other three songs ($3! = 6$ ways), and give the extra song one of the $4$ non-pair labels (empty or a singleton): $4 \cdot 6 \cdot 4 = 96$.

Case 2: one pair-label used twice, the other two once. Choose the repeated label ($3$ ways), choose which two songs carry it ($\binom42 = 6$), and assign the other two labels to the remaining two songs ($2$ ways): $3 \cdot 6 \cdot 2 = 36$.

Total: $96 + 36 = 132$.

The answer is $\boxed{\textbf{(B)}\ 132}$.

Reframing "who likes what" as a label on each song turns a vague social situation into a count of functions from $4$ songs to $7$ labels with a coverage condition. Because three labels must be covered by four songs, the possible multiplicity patterns are just $1$-$1$-$1$-$x$ and $2$-$1$-$1$, and keeping them separate prevents the classic overcount.

Inclusion-exclusion on the missing pair-labels: $7^4 - 3\cdot 6^4 + 3 \cdot 5^4 - 4^4 = 2401 - 3888 + 1875 - 256 = 132$.

• Assigning the three pair-labels to three songs and letting the fourth song take any of 7 labels (4 3! 7 = 168), which double counts the 2-1-1 lists.
• Allowing the fourth song in Case 1 to be liked by all three girls, which adds $24$ forbidden lists.