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 \ | + | Therefore, there are <math>4 \cdot 6 = \boxed{\mathrm{(E) \ } 24}</math> ways to choose the classes. |
==Solution 2== | ==Solution 2== | ||
Line 32: | Line 32: | ||
<math>XOXOXO</math> | <math>XOXOXO</math> | ||
− | There are 6 ways to | + | There are 6 ways to place the first class. |
− | There are 4 ways to | + | There are 4 ways to place the second class. |
− | There is 1 way to | + | There is 1 way to place the third class. |
− | We multiply <math>6 | + | We multiply <math>6 \cdot 4 \cdot 1= \boxed{(E) 24}</math> |
− | + | —Baolan | |
==See Also== | ==See Also== |
Revision as of 21:06, 8 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 1 way to place the third class.
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.