2018 AMC 10A 第 4 题

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

先选出三个互不相邻的课时。

First choose the three nonconsecutive periods

选定课时后，再排列三门不同的课程。

After the periods are chosen, arrange the three named courses

这

3

门数学课可以占据以下课时：

(1, 3, 5)\text{，}

(1, 3, 6)\text{，}

(1, 4, 6)\text{，}

(2, 4, 6)\text{。}

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.

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