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

Revision as of 18:16, 26 March 2018

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

Sup people

Solution 3

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

2018 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 ==Solution 2==
-First draw 6 <math>X</math>'s representing the 6 periods.
-<math>XXXXXX</math>
-Let the <math>O</math>'s represent the classes that occupy each period.
+Sup people
-<math>XOXOXO</math>
-There are 6 ways to place the first class.
-There are 4 ways to place the second class.
-There is 2 way to place the third class.
-But you over counted, so divide by 2.
-We multiply <math>(6 \cdot 4 \cdot 2)/2= \boxed{\textbf{(E) } 24}</math>
-—Baolan
 ==Solution 3==

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