Maths Olympiad Prep

Track / Stage 3 / 62 of 260 #62 of 1964

Problem 62

AMC 10/12, early questions
Combinatorics Difficulty 3.1 Multiple choice

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.)

Pick one

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Official solution

We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods 1,3,51, 3, 5
Periods 1,3,61, 3, 6
Periods 1,4,61, 4, 6
Periods 2,4,62, 4, 6
There are 44 ways to place 33 nondistinguishable classes into 66 periods such that no two classes are in consecutive periods. For each of these ways, there are 3!=63! = 6 orderings of the classes among themselves.
Therefore, there are 46=(E) 244 \cdot 6 = \boxed{\textbf{(E) } 24} ways to choose the classes.
-Versailles15625

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.