2012 AMC 10B Problem 24

There are two cases: Each pair has exactly one liked song in common, or some pair has $2$ liked songs in common. This is because each pair must have at least $1$ liked song in common, and any more pairs than in the cases would result in $5$ songs. Case $1:$ Each pair has exactly one liked song in common There are $4$ ways to choose the song that one pair likes, $3$ ways to choose the song that the second pair likes, and $2$ ways to choose the song the third pair likes if we choose some order for them. Then, for the last song, one of them could like it which has $3$ cases or none of them likes it which is another case. Thus, the number of solutions in this case is $$4\cdot 3\cdot 2\cdot (3+1)=96.$$ Case $2:$ Some pair has $2$ liked songs in common There are $3$ ways to choose the pair that has $2$ liked songs in common. Then, there are $\binom 42 = 6$ ways to choose which two songs they like. Finally, there are $2! =2$ ways to assign the two remaining songs to the other two pairs. Thus, the number of solutions in this case is $$3\cdot 6\cdot 2=36$$ The total amount is then $96+36 = 132.$ Thus, the correct answer is B .