2018 AMC 12A Problem 3

Problem 3 of 25EasierCombinatorics

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.)

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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.

Tagged: arrangements with restrictions · permutations