2018 AMC 12A Problem 3

Below is the professionally curated solution for Problem 3 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:arrangements with restrictions permutations

Difficulty rating: 1130

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.)

$3$

$6$

$12$

$18$

$24$

Solution:

The choices of three non-consecutive periods are $\{1,3,5\},$ $\{1,3,6\},$ $\{1,4,6\},$ and $\{2,4,6\},$ a total of $4.$ The three distinct courses can be placed into any such set in $3! = 6$ orders, giving $4 \cdot 6 = 24$ schedules.

Thus, the correct answer is E.

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