2018 AMC 10A Problem 4

Attempt Problem 4 of the 2018 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

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.)
33
66
1212
1818
2424
Answer: E
Concepts:arrangements with restrictionspermutationssystematic listing
Small hint:
First choose the three nonconsecutive periods
Big hint:
After the periods are chosen, arrange the three named courses
Solution:
The 33 classes can occupy the following periods: (1,3,5),(1, 3, 5),(1,3,6),(1, 3, 6),(1,4,6), (1, 4, 6),(2,4,6). (2, 4, 6).
This means that there are 44 ways to choose which periods the mathematics courses occur.
For each configuration, there are 3!3! ways to determine the order of the courses, for a total of 6⋅4=246 \cdot 4 = 24 schedules.
Thus, E is the correct answer.
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