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Revision as of 21:51, 28 October 2022
- The following problem is from both the 2018 AMC 10A #4 and 2018 AMC 12A #3, so both problems redirect to this page.
Contents
Problem
How many ways can a student schedule 
mathematics courses -- algebra, geometry, and number theory -- in a 
-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 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
Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) 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
Solution 3
We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s:
![\[10101\]](http://latex.artofproblemsolving.com/1/4/6/1468e466217ce2022f0a100b0a1ded18ae553b9f.png)
Now we need to place 1 additional 0. There are 4 places to put it: ![\[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\]](http://latex.artofproblemsolving.com/0/9/e/09e6ed2c1a7825673c807ea0c155b99bb0c0a241.png)
It can be placed in any 1 of the underscores.
Since there are 
ways to order the math classes, the answer is 
.
Video Solution
~IceMatrix
Video Solution
~savannahsolver
Video Solution
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. 
