Question

一个圆被分成 $6$ 个扇形，且这 $6$ 个扇形大小不同。接着把其中 $2$ 个扇形涂红、 $2$ 个涂绿、 $2$ 个涂蓝，并要求相邻的两个扇形颜色不同。下图展示了一种涂色方式。

一共有多少种不同的涂色方式？

A circle has been divided into $6$ sectors of $6$ different sizes. Then $2$ of the sectors are painted red, $2$ painted green, and $2$ painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.

How many different colorings are possible?

$12$

$16$

$18$

$24$

$28$

答案：D

知识点：图论补集计数分类讨论

难度评级：1800

解答：

六个大小不同的扇形构成一个由 $6$ 个可区分位置组成的固定环，所以要计算 $6$ 环的合法 $3$ 色涂色，并要求每种颜色恰好使用两次。合法 $3$ 色涂色共有 $(3-1)^6 + (3-1) = 66$ 种。其中有 $6$ 种只使用两种颜色：缺少的颜色有 $3$ 种选择，其余两种颜色有 $2$ 种交替方式。另有 $36$ 种的颜色计数为 $(3,2,1)$ ：出现三次的颜色有 $3$ 种选择，它所占的交替位置组有 $2$ 种选择，出现两次的颜色有 $2$ 种选择，再从其余 $3$ 个位置中选择它的 $2$ 个位置，有 $3$ 种方法。剩下的 $66 - 6 - 36 = 24$ 种恰好每种颜色使用两次。所以答案是 D。

The six unequal sectors form a fixed cycle of $6$ distinguishable positions, so we want proper $3$ -colorings of a $6$ -cycle that use each color exactly twice. There are $(3-1)^6 + (3-1) = 66$ proper $3$ -colorings altogether. Of these, $6$ use only two colors: choose the missing color in $3$ ways, then alternate the other two colors in $2$ ways. Another $36$ have color counts $(3,2,1)$ : choose the color used three times in $3$ ways, choose one of the $2$ alternating sets of positions for it, choose the color used twice in $2$ ways, and choose its $2$ positions among the other $3$ in $3$ ways. The remaining $66 - 6 - 36 = 24$ use each color exactly twice. Therefore, the answer is D.

2025 AMC 10B 第 16 题

答案：D

难度评级：1800

解答：

其他年份的第 16 题