Question

一天有 $6$ 节课。若任意两门数学课都不能安排在相邻课时，那么一名学生安排 $3$ 门数学课，包括代数、几何和数论，共有多少种排法？

（其余 $3$ 节课所上的课程无需考虑。）

How many ways can a student schedule $3$ mathematics courses — algebra, geometry, and number theory — in a $6$ -period day if no two mathematics courses can be taken in consecutive periods?

(What courses the student takes during the other $3$ periods is of no concern here.)

$3$

$6$

$12$

$18$

$24$

答案：E

知识点：有限制的排列排列系统列举

难度评级：1220

解答：

这 $3$ 门数学课可以占据以下课时： $(1, 3, 5),$ $(1, 3, 6),$ $(1, 4, 6),$ $(2, 4, 6).$

因此，选择数学课所在课时共有 $4$ 种方法。

每种课时安排中，三门课有 $3!$ 种顺序，所以总共有 $6 \cdot 4 = 24$ 种课程表。

因此正确答案是 E。

The $3$ classes can occupy the following periods: $(1, 3, 5),$ $(1, 3, 6),$ $(1, 4, 6),$ $(2, 4, 6).$

This means that there are $4$ ways to choose which periods the mathematics courses occur.

For each configuration, there are $3!$ ways to determine the order of the courses, for a total of $6 \cdot 4 = 24$ schedules.

Thus, E is the correct answer.

2018 AMC 10A 第 4 题

答案：E

难度评级：1220

解答：

其他年份的第 4 题