Difference between revisions of "2018 AMC 10A Problems/Problem 4"
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There are <math>4</math> ways to place <math>3</math> nondistinguishable classes into <math>6</math> periods such that no two classes are in consecutive periods. For each of these ways, there are <math>3! = 6</math> orderings of the classes among themselves. | There are <math>4</math> ways to place <math>3</math> nondistinguishable classes into <math>6</math> periods such that no two classes are in consecutive periods. For each of these ways, there are <math>3! = 6</math> orderings of the classes among themselves. | ||
| − | Therefore, there are <math>4 \cdot 6 = \boxed{\ | + | Therefore, there are <math>4 \cdot 6 = \boxed{\textbf{(E) \ } 24}</math> ways to choose the classes. |
==Solution 2== | ==Solution 2== | ||
Revision as of 19:49, 9 February 2018
Contents
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 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.
Solution 2
First draw 6 
's representing the 6 periods.
Let the 
's represent the classes that occupy each period.
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 
—Baolan
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. 
