# Difference between revisions of "2018 AMC 10A Problems/Problem 4"

## Problem

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

## Solution 1

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

## Solution 2

Realize that the number of ways of placing, regardless of order, the 3 mathematics courses in a 6-period day so that no two are consecutive is the same as the number of ways of placing 3 mathematics courses in a sequence of 4 periods regardless of order and whether or not they are consecutive.

To see that there is a one to one correlation, note that for every way of placing 3 mathematics courses in 4 total periods (as above) one can add a non-mathematics course between each pair (2 total) of consecutively occurring mathematics courses (not necessarily back to back) to ensure there will be no two consecutive mathematics courses in the resulting 6-period day. For example, where $M$ denotes a math course and $O$ denotes a non-math course: $M O M M \rightarrow M O O M O M$

For each 6-period sequence consisting of $M$s and $O$s, we have $3!$ orderings of the 3 distinct mathematics courses.

So, our answer is $\dbinom{4}{3}(3!)= \boxed{\textbf{(E) } 24}$

- Gregwwl

## Solution 3

Counting what we don't want is another slick way to solve this problem. Use PIE to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.

Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$. Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.

Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$. Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.

By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$. Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{\textbf{(E) } 24}$

-LitJamal

## Video Solution

~savannahsolver

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