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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>.
 
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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==Solution 2==
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Denote the <math>6</math> discs as in the first solution. Ignoring reflections or rotations, there are <math>\binom{6}{3} * \binom{3}{2} = 60</math> colorings. Now we need to count the number of fixed points under possible transformations:
 +
 +
1. The identify transformation. Since this doesn't change anything, there are <math>60</math> fixed points
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2. Reflect about a line of symmetry. There are <math>3</math> lines of reflections. Take the line of reflection going through the centers of circles <math>1</math> and <math>5</math>. Then, the colors of circles <math>2</math> and <math>3</math> must be the same, and the colors of circles <math>4</math> and <math>6</math> must be the same. This gives us <math>4</math> fixed points per line of reflection
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3. Rotate by <math>120^\circ</math> counter clockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles <math>1</math>, <math>4</math>, and <math>6</math> will be the same. Similarly, the colors of circles <math>2</math>, <math>3</math>, and <math>5</math> will be the same. This is impossible, so this case gives us <math>0</math> fixed points per rotation.
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By Burnside's lemma, the total number of colorings is <math>(1*60+3*4+2*0)/(1+3+2) = \boxed{\textbf{(D) } 12}</math>.
  
 
==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 19:03, 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?

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

Solution

First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$. For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}$.

Solution 2

Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} * \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:

1. The identify transformation. Since this doesn't change anything, there are $60$ fixed points

2. Reflect about a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$. Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection

3. Rotate by $120^\circ$ counter clockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$, $4$, and $6$ will be the same. Similarly, the colors of circles $2$, $3$, and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.

By Burnside's lemma, the total number of colorings is $(1*60+3*4+2*0)/(1+3+2) = \boxed{\textbf{(D) } 12}$.

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

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