2013 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 of the 2013 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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Concepts:caseworkmultiplication principle

Difficulty rating: 1880

15.

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cottontail. 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?

9696

108108

156156

204204

372372

Solution:

If the two parents share a store, there are 44 choices for it, and each child must go to one of the other three stores: 433=1084\cdot 3^3 = 108 ways.

If the parents go to different stores, there are 43=124\cdot 3 = 12 choices, and each child must go to one of the two remaining stores: 1223=9612\cdot 2^3 = 96 ways.

The total is 108+96=204.108 + 96 = 204.

Thus, the correct answer is D.

Problem 15 in Other Years