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Difference between revisions of "2013 AMC 12A Problems/Problem 15"

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== Problem==
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== Problem ==
  
 
Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
 
Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
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<math>\textbf{(A)} \ 96 \qquad  \textbf{(B)} \ 108 \qquad  \textbf{(C)} \ 156 \qquad  \textbf{(D)} \ 204 \qquad  \textbf{(E)} \ 372 </math>
 
<math>\textbf{(A)} \ 96 \qquad  \textbf{(B)} \ 108 \qquad  \textbf{(C)} \ 156 \qquad  \textbf{(D)} \ 204 \qquad  \textbf{(E)} \ 372 </math>
  
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==Solution 1==
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== Solution 1 ==
  
 
There are two possibilities regarding the parents.
 
There are two possibilities regarding the parents.
  
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1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are <math>4 * 3^3 = 108</math> combinations.
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1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are <math>4 \cdot 3^3 = 108</math> combinations.
  
  
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2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are <math>4 * 3 * 2^3 = 96</math> combinations.
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2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are <math>4 \cdot 3 \cdot 2^3 = 96</math> combinations.
  
 
Adding up, we get <math>108 + 96 = 204</math> combinations.
 
Adding up, we get <math>108 + 96 = 204</math> combinations.
  
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==Solution 2==
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== Solution 2 ==
  
 
We tackle the problem by sorting it by how many stores are involved in the transaction.
 
We tackle the problem by sorting it by how many stores are involved in the transaction.
  
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1) 2 stores are involved. There are <math>\binom{4}{2} = 6</math> ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. <math>6 * 2 = 12 total arrangements</math>.
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1) 2 stores are involved. There are <math>\binom{4}{2} = 6</math> ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. <math>6 \cdot 2 = 12</math> total arrangements.
  
  
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'''Together:'''
 
'''Together:'''
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Both parents are in one store and the 3 children are split between the other two. There are <math>\binom{3}{2}</math> ways to split the children and <math>3!</math> ways to choose to which store each group will be sold. <math>3! * \binom{3}{2} = 18</math>.
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Both parents are in one store and the 3 children are split between the other two. There are <math>\binom{3}{2}</math> ways to split the children and <math>3!</math> ways to choose to which store each group will be sold. <math>3! \cdot \binom{3}{2} = 18</math>.
  
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<math>(6 + 18) * 4 = 96</math> total arrangements.
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<math>(6 + 18) \cdot 4 = 96</math> total arrangements.
  
  
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'''Separately:'''
 
'''Separately:'''
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All children must be split between two stores. There are <math>\binom{3}{2} = 3</math> ways to arrange this. We can then arrange which group is sold to which store in <math>4!</math> ways. <math>4! * 3 = 72</math>.
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All children must be split between two stores. There are <math>\binom{3}{2} = 3</math> ways to arrange this. We can then arrange which group is sold to which store in <math>4!</math> ways. <math>4! \cdot 3 = 72</math>.
  
 
'''Together:'''
 
'''Together:'''
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'''Final Answer:'''
 
'''Final Answer:'''
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<math>12 + 96 + 96 = 204</math>
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<math>12 + 96 + 96 = \boxed{\textbf{(D)} \: 204}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:02, 8 November 2024

Problem

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?

$\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372$

Solution 1

There are two possibilities regarding the parents.

1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are $4 \cdot 3^3 = 108$ combinations.


2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are $4 \cdot 3 \cdot 2^3 = 96$ combinations.

Adding up, we get $108 + 96 = 204$ combinations.

Solution 2

We tackle the problem by sorting it by how many stores are involved in the transaction.

1) 2 stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.


2) 3 stores are involved. There are $\binom{4}{3} = 4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.


Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total.

Together: Both parents are in one store and the 3 children are split between the other two. There are $\binom{3}{2}$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3! \cdot \binom{3}{2} = 18$.

$(6 + 18) \cdot 4 = 96$ total arrangements.


3) All 4 stores are involved. We break down the problem as previously shown.


Separately: All children must be split between two stores. There are $\binom{3}{2} = 3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4! \cdot 3 = 72$.

Together: Both parents are in one store and the 3 children are each in another store. There are $4! = 24$ ways to arrange this.

$24 + 72 = 96$ total arrangements.


Final Answer: $12 + 96 + 96 = \boxed{\textbf{(D)} \: 204}$.

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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