Question

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.)

$3$

$6$

$12$

$18$

$24$

Answer: E

Concepts:arrangements with restrictions permutations systematic listing

Difficulty rating: 1220

Solution:

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 Problem 4

Answer: E

Difficulty rating: 1220

Solution:

Problem 4 in Other Years