Difference between revisions of "2012 AMC 12B Problems/Problem 16"
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Revision as of 11:06, 4 July 2013
- The following problem is from both the 2012 AMC 12B #16 and 2012 AMC 10B #24, so both problems redirect to this page.
Problem
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?
Solution
Let the ordered triple denote that songs are liked by Amy and Beth, songs by Beth and Jo, and songs by Jo and Amy. We claim that the only possible triples are .
To show this, observe these are all valid conditions. Second, note that none of can be bigger than 3. Suppose otherwise, that . 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 or to be at least 1. In fact, we require either or to equal 1, otherwise there will be a song liked by all three. Suppose . Then we must have since no song is liked by all three girls, a contradiction.
Case 1:How many ways are there for to equal ? 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 in ways.
Case 2:To find the number of ways for , observe there are 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 let any three pairs of the girls like two songs, so we multiply by 3. In this case there are ways for the girls to like the songs.
That gives a total of ways for the girls to like the song, so the answer is .
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.