Question

在下图中， $6$ 个圆盘中有 $3$ 个要涂成蓝色， $2$ 个要涂成红色， $1$ 个要涂成绿色。如果两个涂色方案可以通过整个图形的旋转或反射相互得到，则视为相同。共有多少种不同涂法？

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：组合对称性分类讨论

难度评级：2010

解答：

先固定绿色圆盘在最上方。这个位置与其他外层位置旋转等价。绿色圆盘在内层 $\binom52=10$ 个位置的情形数相同，因此最后将外层情形数乘以。

剩下五个位置中选两个放红色，原本有 $8$ 种；在保留绿色位置的反射对称下，其中 $2$ 种自身对称，其余成对等价，所以不同放法数为 $4$ ，再加上这 $2+4=6$ 种，共种。

因此总数为 $6+6=12$ 。

所以正确答案是 D。

By symmetry, the green disk has two possible types of position: a corner or a side midpoint. Fix one representative of either type. There are $\binom52=10$ ways to choose the two red disks.

The reflection that fixes the green position fixes one of the other disks and exchanges the other four disks in two pairs. Exactly $2$ red-disk choices are unchanged by this reflection: choosing either exchanged pair. The other $8$ choices form $4$ mirror-image pairs. Hence there are $2+4=6$ paintings for each type of green position.

The two types therefore give $6+6=12$ paintings.

Thus, the correct answer is D .

2017 AMC 10B 第 18 题

答案：D

难度评级：2010

解答：

其他年份的第 18 题