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2018 AMC 12A Problem 3

Problem 3 of 25EasierCounting & Probability

How many ways can a student schedule 33 mathematics courses—algebra, geometry, and number theory—in a 66-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 33 periods is of no concern here.)

Answer choices

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Solution

The choices of three non-consecutive periods are {1,3,5},\{1,3,5\}, {1,3,6},\{1,3,6\}, {1,4,6},\{1,4,6\}, and {2,4,6},\{2,4,6\}, a total of 4.4. The three distinct courses can be placed into any such set in 3!=63! = 6 orders, giving 46=244 \cdot 6 = 24 schedules. Thus, the correct answer is E.

More practice

Concepts: arrangements with restrictions · permutations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.