Difference between revisions of "2013 AMC 12A Problems/Problem 15"
Epicwisdom (talk | contribs) (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....") |
|||
| Line 1: | Line 1: | ||
| + | == 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? | ||
| + | |||
| + | <math>\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
There are two possibilities regarding the parents. | There are two possibilities regarding the parents. | ||
| Line 6: | Line 14: | ||
Adding up, we get <math>108 + 96 = 204</math> combinations. | Adding up, we get <math>108 + 96 = 204</math> combinations. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=14|num-a=16}} | ||
Revision as of 18:42, 22 February 2013
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
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.
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 | |
