Difference between revisions of "2018 AMC 10A Problems/Problem 4"
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+ | {{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #3]] and [[2018 AMC 10A Problems|2018 AMC 10A #4]]}} | ||
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==Problem== | ==Problem== | ||
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<math>\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24</math> | <math>\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24</math> | ||
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==Solution 1== | ==Solution 1== | ||
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-LitJamal | -LitJamal | ||
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+ | ==Video Solutions== | ||
+ | https://youtu.be/vO-ELYmgRI8 | ||
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+ | https://youtu.be/kBOCP8p8-o8 | ||
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+ | ~savannahsolver | ||
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+ | https://youtu.be/GHCecqsF8es | ||
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+ | Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2018|ab=A|num-b=3|num-a=5}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 17:53, 30 December 2020
- The following problem is from both the 2018 AMC 12A #3 and 2018 AMC 10A #4, so both problems redirect to this page.
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.)
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
Periods
Periods
Periods
There are ways to place nondistinguishable classes into periods such that no two classes are in consecutive periods. For each of these ways, there are orderings of the classes among themselves.
Therefore, there are 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 denotes a math course and denotes a non-math course:
For each 6-period sequence consisting of s and s, we have orderings of the 3 distinct mathematics courses.
So, our answer is
- 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, . Finally we have 4 periods remaining to place the final math class. Thus there are 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, . Thus there are ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: . Then we subtract that answer from the total ways to place the classes with no restrictions:
-LitJamal
Video Solutions
~savannahsolver
Education, the Study of Everything
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.