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

(Created page with "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....")
 
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== Problem ==
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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>
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== Solution 1 ==
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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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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 \cdot 2 = 12</math> total arrangements.
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2) 3 stores are involved. There are <math>\binom{4}{3} = 4</math> 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.
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'''Separately:'''
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All children must be in one store. There are <math>3!</math> ways to arrange this. <math>6</math> ways in total.
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'''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! \cdot \binom{3}{2} = 18</math>.
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<math>(6 + 18) \cdot 4 = 96</math> total arrangements.
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3) All 4 stores are involved. We break down the problem as previously shown.
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'''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! \cdot 3 = 72</math>.
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'''Together:'''
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Both parents are in one store and the 3 children are each in another store. There are <math>4! = 24</math> ways to arrange this.
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<math>24 + 72 = 96</math> total arrangements.
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'''Final Answer:'''
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<math>12 + 96 + 96 = \boxed{\textbf{(D)} \: 204}</math>.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=14|num-a=16}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

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
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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