AMC12 2017 B

AMC12 2017 B · Q13

AMC12 2017 B · Q13. It mainly tests Casework, Counting with symmetry / Burnside (rare).

In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

在下面的图形中，要将 6 个圆盘中的 3 个涂成蓝色，2 个涂成红色，1 个涂成绿色。通过整个图形的旋转或反射可以相互得到的两种涂法视为相同。有多少种不同的涂法可能？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): By symmetry, there are just two cases for the position of the green disk: corner or non-corner. If a corner disk is painted green, then there is 1 case in which both red disks are adjacent to the green disk, there are 2 cases in which neither red disk is adjacent to the green disk, and there are 3 cases in which exactly one of the red disks is adjacent to the green disk. Similarly, if a non-corner disk is painted green, then there is 1 case in which neither red disk is in a corner, there are 2 cases in which both red disks are in a corner, and there are 3 cases in which exactly one of the red disks is in a corner. The total number of paintings is $1 + 2 + 3 + 1 + 2 + 3 = 12$.

答案（D）：由于对称性，绿色圆片的位置只有两种情况：在角上或不在角上。若角上的圆片被涂成绿色，则有 1 种情况两个红色圆片都与绿色圆片相邻，有 2 种情况两个红色圆片都不与绿色圆片相邻，还有 3 种情况恰好有一个红色圆片与绿色圆片相邻。类似地，若一个非角上的圆片被涂成绿色，则有 1 种情况两个红色圆片都不在角上，有 2 种情况两个红色圆片都在角上，还有 3 种情况恰好有一个红色圆片在角上。涂色方案总数为 $1 + 2 + 3 + 1 + 2 + 3 = 12$。

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