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>
  
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==Solutions==
  
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=== Solution 1===
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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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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.
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'''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.
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'''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.
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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>.
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=== Solution 2===
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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>.
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Case 1: Fourth song = <math>N, A, B, J </math>
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Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.
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Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.
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Case 2: Fourth song = <math>AB, BJ, AJ</math>
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Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.
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Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.
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<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.
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=== Solution 3===
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There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.
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There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.
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In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.
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There are <math>3</math> cases for the 4th song, call it song D.
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Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.
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Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.
  
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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.
  
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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.
  
== Solution==
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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>
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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~ Nafer
  
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.
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==Video Solutions:==
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2012amc10b/272
  
'''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.
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~dolphin7
  
'''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 = \boxed{132}</math> ways for the girls to like the song, so the answer is <math>(\textrm{\textbf{B}})</math>.
 
  
 
== See Also ==
 
== See Also ==
  
  
{{AMC10 box|year=2012|ab=B|num-b=23|after=25}}
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{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}
  
 
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

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.

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?

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

Solutions

Solution 1

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 $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\boxed{(\textrm{\textbf{B}})}$.

Solution 2

Let $AB, BJ$, and $AJ$ 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 $A, B, J,$ and $N$ 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 $1\: AB, BJ$, and $AJ$, they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$, and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$, and a song from the set $\{N, A, B, J, AB, BJ, AJ\}$.

Case 1: Fourth song = $N, A, B, J$

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 = $(4!)$ rearrangements for each choice $*\: 4$ choices = $96$.

Case 2: Fourth song = $AB, BJ, AJ$

Note that in Case $2$, all three of the choices for the fourth song repeat somewhere in the first three songs.

Number of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\: 3$ choices = $36$.

$96 + 36 = \boxed{\textbf{(B)} \: 132}$.

Solution 3

There are $\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.

There are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.

In total, there are $\binom{4}{3}\cdot3!$ possibilities for the first $3$ songs.

There are $3$ cases for the 4th song, call it song D.

Case $1$: D is disliked by all $3$ girls $\implies$ there is only $1$ possibility.

Case $2$: D is liked by exactly $1$ girl $\implies$ there are $3$ possibility.

Case $3$: D is liked by exactly $2$ girls $\implies$ there are $3$ 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 $3$ songs liked by the same girls.

Counting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\cdot12=36$ overlapped possibilities.

Finally, there are $\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\boxed{\mathrm{(B)}}$

~ Nafer

Video Solutions:

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc10b/272

~dolphin7


See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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