Difference between revisions of "2017 AMC 10B Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | + | First we figure out the number of ways to put the <math>3</math> blue disks. Denote the spots to put the disks as <math>1-6</math> from left to right, top to bottom. The cases to put the blue disks are <math>(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)</math>. For each of those cases we can easily figure out the number of ways for each case, so the total amount is <math>2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}</math>. | |
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2017|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:30, 16 February 2017
Problem
In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
Solution
First we figure out the number of ways to put the 
blue disks. Denote the spots to put the disks as 
from left to right, top to bottom. The cases to put the blue disks are 
. For each of those cases we can easily figure out the number of ways for each case, so the total amount is 
.
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
