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Difference between revisions of "2012 AMC 12B Problems/Problem 16"
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== Solution==
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== Solution One==
 
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>.  
 
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>.  
  
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That gives a total of <math>96 + 36 = \boxed{132}</math> ways for the girls to like the song, so the answer is <math>(\textrm{\textbf{B}})</math>.
 
That gives a total of <math>96 + 36 = \boxed{132}</math> ways for the girls to like the song, so the answer is <math>(\textrm{\textbf{B}})</math>.
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== Solution Two==
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We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are <math>2\cdot2\cdot2=8</math> ways to assign song preferences for the fourth song. Multiplying, we obtain an answer of <math>4\cdot3\cdot2\cdot8=192</math>. However, in doing so, we have committed an egregious error. We have in fact over counted the cases in which the fourth song is liked by two girls but not the third.
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We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are <math>4\cdot3\cdot2\cdot5=120</math> of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for!
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In performing our past two calculations, we have, however, established that the answer has a lower bound of <math>120</math> and an upper bound of <math>192</math>. As <math>(\textrm{\textbf{B}})</math> is the only answer within these bounds, we conclude that the answer must be <math>(\textrm{\textbf{B}})</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 23:55, 17 January 2014

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?

$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105$



Solution One

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. We claim that the only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$.

To show this, 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, a contradiction.

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 $(a,b,c) = (2,1,1)$, 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 let any three pairs of the girls 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 $96 + 36 = \boxed{132}$ ways for the girls to like the song, so the answer is $(\textrm{\textbf{B}})$.

Solution Two

We begin by noticing that there are four ways to assign a song liked by both Amy and Beth, three ways to assign a song liked by both Amy and Jo (because Jo may not like the song liked by both Amy and Beth), and two ways to assign a song liked by both Beth and Jo (because both Beth and Jo may not like the song liked by the previous pairs. Additionally, there are $2\cdot2\cdot2=8$ ways to assign song preferences for the fourth song. Multiplying, we obtain an answer of $4\cdot3\cdot2\cdot8=192$. However, in doing so, we have committed an egregious error. We have in fact over counted the cases in which the fourth song is liked by two girls but not the third.

We proceed again by ignoring the cases in which the fourth song is liked by two girls but not the third. There are $4\cdot3\cdot2\cdot5=120$ of these cases. However, in doing so, we have committed yet another egregious error. The cases in which the fourth song is liked by two girls but not the third have not been accounted for!

In performing our past two calculations, we have, however, established that the answer has a lower bound of $120$ and an upper bound of $192$. As $(\textrm{\textbf{B}})$ is the only answer within these bounds, we conclude that the answer must be $(\textrm{\textbf{B}})$.

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

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