AMC12 2012 B

Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$. To show this, first observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$. Then we must have $c=0$ since no song is liked by all three girls, which contradicts with the statement "for each of the three pairs of the girls, there is at least one song liked by those two girls". Case 1: How many ways are there for $(a,b,c)$ to equal $(1,1,1)$? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\cdot3\cdot2\cdot4 = 96$ ways. Case 2: To find the number of ways for $(a,b,c) = (2,1,1)$ (without order), observe there are $\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we can let any three pairs of the girls to like two songs, so we multiply by 3. In this case there are $(6\cdot2)\cdot3=36$ ways for the girls to like the songs. That gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\boxed{(\textrm{\textbf{B}})}$.