AMC12 2013 A

AMC12 2013 A · Q15

AMC12 2013 A · Q15. It mainly tests Basic counting (rules of product/sum), Casework.

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?

兔子 Peter 和 Pauline 有三个后代——Flopsie、Mopsie 和 Cottontail。这五只兔子要分送到四个不同的宠物店，使得没有商店同时得到父母和子女。不要求每个商店都得到兔子。可以有多少种不同的方式？

(A) 96 96

(B) 108 108

(D) 204 204

(E) 372 372

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): There are two cases. If Peter and Pauline are given to the same pet store, then there are 4 ways to choose that store. Each of the children must then be assigned to one of the other three stores, and this can be done in $3^3 = 27$ ways. Therefore there are $4 \cdot 27 = 108$ possible assignments in this case. If Peter and Pauline are given to different stores, then there are $4 \cdot 3 = 12$ ways to choose those stores. In this case, each of the children must be assigned to one of the other two stores, and this can be done in $2^3 = 8$ ways. Therefore there are $12 \cdot 8 = 96$ possible assignments in this case. The total number of assignments is $108 + 96 = 204$.

答案（D）：有两种情况。如果 Peter 和 Pauline 被分配到同一家宠物店，那么选择该店有 4 种方式。然后其余每个孩子必须被分配到另外三家店中的一家，这可以用 $3^3 = 27$ 种方式完成。因此这种情况下共有 $4 \cdot 27 = 108$ 种可能的分配。如果 Peter 和 Pauline 被分配到不同的店，那么选择这两家店有 $4 \cdot 3 = 12$ 种方式。在这种情况下，每个孩子必须被分配到另外两家店中的一家，这可以用 $2^3 = 8$ 种方式完成。因此这种情况下共有 $12 \cdot 8 = 96$ 种可能的分配。分配总数为 $108 + 96 = 204$。

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