Difference between revisions of "2013 AMC 12A Problems/Problem 15"
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| − | == Problem== | + | == 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> | ||
| − | ==Solution 1== | + | == Solution 1 == |
There are two possibilities regarding the parents. | There are two possibilities regarding the parents. | ||
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Adding up, we get <math>108 + 96 = 204</math> combinations. | Adding up, we get <math>108 + 96 = 204</math> combinations. | ||
| − | ==Solution 2== | + | == 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. | ||
Revision as of 14:38, 7 June 2018
Contents
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?
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 
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 
combinations.
Adding up, we get 
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 
ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. 
total arrangements.
2) 3 stores are involved. There are 
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 
ways to arrange this. 
ways in total.
Together:
Both parents are in one store and the 3 children are split between the other two. There are 
ways to split the children and ways to choose to which store each group will be sold. 
.
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 
ways to arrange this. We can then arrange which group is sold to which store in 
ways. 
.
Together:
Both parents are in one store and the 3 children are each in another store. There are 
ways to arrange this.
total arrangements.
Final Answer:
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
