2012 AMC 10B Problem 24

Below is the professionally curated solution for Problem 24 of the 2012 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10B solutions, or check the answer key.

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Concepts:basic countingcaseworkmultiplication principle

Difficulty rating: 2100

24.

Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?

108 108

132 132

671 671

846 846

1105 1105

Solution:

There are two cases: Each pair has exactly one liked song in common, or some pair has 22 liked songs in common. This is because each pair must have at least 11 liked song in common, and any more pairs than in the cases would result in 55 songs.

Case 1: Each pair has exactly one liked song in common

There are 44 ways to choose the song that one pair likes, 33 ways to choose the song that the second pair likes, and 22 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 33 cases or none of them likes it which is another case. Thus, the number of solutions in this case is 432(3+1)=96.4\cdot 3\cdot 2\cdot (3+1)=96.

Case 2: Some pair has 22 liked songs in common

There are 33 ways to choose the pair that has 22 liked songs in common. Then, there are (43)=6\binom 43 = 6 ways to choose which songs they like. Finally, there are (21)=2\binom 21 =2 ways to figure out who likes the last song. Thus, the number of solutions in this case is 362=363\cdot 6\cdot 2=36

The total amount is then 96+36=132.96+36 = 132.

Thus, the correct answer is B .

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