Question

如下图所示，要把 $6$ 个圆盘中的 $3$ 个涂成蓝色， $2$ 个涂成红色， $1$ 个涂成绿色。如果两种涂色可以通过整个图形的旋转或翻折互相得到，则认为它们相同。共有多少种不同的涂色？

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?

$6$

$8$

$9$

$12$

$15$

答案：D

知识点：分类讨论对称性

难度评级：1660

解答：

该图形有 $3$ 个角上的圆盘和 $\dfrac{6!}{3!2!}=60$ 个非角上的圆盘，对称群与三角形相同。 $3,2,1$ 。

固定绿色圆盘的类型。 $3$ $2$ $4$ $2$ $3$ $2$ $2\cdot2=4$ $\dfrac{60+3(4)}{6}=12$

如果绿色在角上，两个红色圆盘可以都与绿色相邻、恰有一个与绿色相邻、或都不相邻，给出种不同涂色。如果绿色不在角上，两个红色圆盘可以有两个、一个或零个在角上，同样得到种涂色。蓝色圆盘填满其余位置，所以总数为。所以正确答案是 D。

Before accounting for symmetry, there are $\dfrac{6!}{3!2!}=60$ paintings. The two nonidentity rotations partition the disks into two $3$ -cycles, so neither can fix a painting having color counts $3,2,1.$

Each of the $3$ reflections fixes $2$ disks and swaps the other $4$ in $2$ pairs. For a painting to be fixed, the lone green disk and one of the $3$ blue disks must occupy the two fixed positions, in $2$ orders. Of the two swapped pairs, either one can be the red pair, giving $2\cdot2=4$ fixed paintings per reflection. Burnside's Lemma therefore gives $\dfrac{60+3(4)}{6}=12$ distinct paintings.

Thus, the correct answer is D.

2017 AMC 12B 第 13 题

答案：D

难度评级：1660

解答：

其他年份的第 13 题