AMC10 2017 B

AMC10 2017 B · Q18

AMC10 2017 B · Q18. 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?

在下图中，6个圆盘中有3个涂蓝色，2个涂红色，1个涂绿色。通过整个图形的旋转或反射可以得到的两种涂色视为相同。有多少种不同的涂色可能？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

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.

由对称性，绿色圆盘位置有两种情况：角或非角。如果角圆盘涂绿色，则两种红色圆盘都邻接绿色的有1种，两种都不邻接的有2种，正好一种邻接的有3种。类似地，如果非角圆盘涂绿色，则两种红色都不在角的有1种，都在角的有2种，正好一种在角的有3种。总涂色数为1+2+3+1+2+3=12。

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