Difference between revisions of "2012 AMC 12B Problems/Problem 16"
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<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math> | <math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math> | ||
+ | ==Solutions== | ||
+ | === Solution 1=== | ||
+ | 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>. | ||
+ | |||
+ | To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. 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 <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction. | ||
+ | |||
+ | '''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? 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 <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways. | ||
+ | |||
+ | '''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> 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 <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs. | ||
+ | |||
+ | That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>. | ||
+ | |||
+ | === Solution 2=== | ||
+ | |||
+ | |||
+ | Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>. | ||
+ | |||
+ | Case 1: Fourth song = <math>N, A, B, J </math> | ||
+ | Note that in Case 1, all four of the choices for the fourth song are different from the first three songs. | ||
+ | Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>. | ||
− | == | + | Case 2: Fourth song = <math>AB, BJ, AJ</math> |
− | + | ||
+ | Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs. | ||
+ | |||
+ | Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>. | ||
+ | |||
+ | <math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>. | ||
+ | |||
+ | === Solution 3=== | ||
+ | |||
+ | |||
+ | There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls. | ||
+ | |||
+ | There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls. | ||
+ | |||
+ | In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs. | ||
+ | |||
+ | There are <math>3</math> cases for the 4th song, call it song D. | ||
+ | |||
+ | Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility. | ||
+ | |||
+ | Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility. | ||
+ | |||
+ | Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls. | ||
+ | |||
+ | Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities. | ||
+ | |||
+ | Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math> | ||
+ | |||
+ | ~ Nafer | ||
− | + | ==Video Solutions:== | |
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc10b/272 | ||
− | + | ~dolphin7 | |
− | |||
− | |||
== See Also == | == See Also == |
Latest revision as of 20:42, 16 May 2021
- The following problem is from both the 2012 AMC 12B #16 and 2012 AMC 10B #24, so both problems redirect to this page.
Contents
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?
Solutions
Solution 1
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 songs, so the answer is .
Solution 2
Let , and denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let and denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least , and , they must be songs out of the that Amy, Beth, and Jo listened to. The fourth song can be of any type , and (there is no because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange , and a song from the set .
Case 1: Fourth song =
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.
Number of ways to rearrange = rearrangements for each choice choices = .
Case 2: Fourth song =
Note that in Case , all three of the choices for the fourth song repeat somewhere in the first three songs.
Number of ways to rearrange = rearrangements for each choice choices = .
.
Solution 3
There are ways to choose the three songs that are liked by the three pairs of girls.
There are ways to determine how the three songs are liked, or which song is liked by which pair of girls.
In total, there are possibilities for the first songs.
There are cases for the 4th song, call it song D.
Case : D is disliked by all girls there is only possibility.
Case : D is liked by exactly girl there are possibility.
Case : D is liked by exactly girls there are pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first songs liked by the same girls.
Counting the overlaps, there are ways to choose the pair with overlaps and ways to choose what the other pairs like independently. In total, there are overlapped possibilities.
Finally, there are ways for the songs to be likely by the girls.
~ Nafer
Video Solutions:
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10b/272
~dolphin7
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.