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Difference between revisions of "2017 AMC 12B Problems/Problem 13"

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==Solution==
 
==Solution==
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Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green ball is in a corner. This yields <math>6</math> possible arrangements for the <math>3</math> blue balls and <math>2</math> red balls in the remaining available slots. Now, consider the case that the green ball is on an edge. This yields <math>6</math> more possible arrangements for the <math>3</math> blue balls and <math>2</math> red balls in the remaining available slots. Thus, our answer is <math>6 + 6 = \boxed{\bold{(D)}\, 12}</math>
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Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields <math>6</math> possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields <math>6</math> more possible arrangements for the <math>3</math> blue disks and <math>2</math> red disks in the remaining available slots. Thus, our answer is <math>6 + 6 = \boxed{\bold{(D)}\, 12}</math>
  
 
Solution by akaashp11
 
Solution by akaashp11

Revision as of 18:48, 19 February 2017

Problem 13

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?

[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]

$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Solution

Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields $6$ more possible arrangements for the $3$ blue disks and $2$ red disks in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{\bold{(D)}\, 12}$

Solution by akaashp11

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

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