Difference between revisions of "2013 AMC 12A Problems/Problem 15"
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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. | ||
− | 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 | + | 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:''' | ||
− | 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! | + | 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>. |
− | <math>(6 + 18) | + | <math>(6 + 18) \cdot 4 = 96</math> total arrangements. |
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'''Separately:''' | '''Separately:''' | ||
− | 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! | + | 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:''' |
Revision as of 21:03, 14 February 2017
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.