Difference between revisions of "2018 AMC 10A Problems/Problem 4"
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Therefore, there are <math>4 \times 6 = \boxed{\mathrm{(E) \ } 24}</math> ways to choose the classes. | Therefore, there are <math>4 \times 6 = \boxed{\mathrm{(E) \ } 24}</math> ways to choose the classes. | ||
| − | ==See | + | ==See also== |
| − | {{AMC10 box|year= | + | {{AMC10 box|year=2018|ab=A|before=[[2017 AMC 10B]]|after=[[2018 AMC 10B]]}} |
| + | * [[AMC 10]] | ||
| + | * [[AMC 10 Problems and Solutions]] | ||
| + | * [[Mathematics competitions]] | ||
| + | * [[Mathematics competition resources]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:12, 8 February 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.)
Solution
We must place the classes into the periods such that no two balls are in the same period or in consecutive period.
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.
See also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by 2017 AMC 10B |
Followed by 2018 AMC 10B | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
