2013 AMC 12A Problems/Problem 15

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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 $4 * 3^3 = 108$ 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 $4 * 3 * 2^3 = 96$ combinations.

Adding up, we get $108 + 96 = 204$ combinations.

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