Track / Stage 3 / 50 of 260 #50 of 2000

Problem 50

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Find the answer

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.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Your answer

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 $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{\textbf{(E) } 24}$ ways to choose the classes.
-Versailles15625

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.