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Difference between revisions of "2022 AMC 10A Problems/Problem 9"
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A rectangle is partitioned into 5 regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
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{{duplicate|[[2022 AMC 10A Problems/Problem 9|2022 AMC 10A #9]] and [[2022 AMC 12A Problems/Problem 7|2022 AMC 12A #7]]}}
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<asy>
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size(5.5cm);
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==Problem==
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draw((0,0)--(0,2)--(2,2)--(2,0)--cycle);
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draw((2,0)--(8,0)--(8,2)--(2,2)--cycle);
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A rectangle is partitioned into <math>5</math> regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
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draw((8,0)--(12,0)--(12,2)--(8,2)--cycle);
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draw((0,2)--(6,2)--(6,4)--(0,4)--cycle);
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<asy> size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); </asy>
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draw((6,2)--(12,2)--(12,4)--(6,4)--cycle);
 
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</asy>
 
  
 
<math>\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720</math>
 
<math>\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720</math>
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==Solution 1==
 
==Solution 1==
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The top left rectangle can be <math>5</math> possible colors. Then the bottom left region can only be <math>4</math> possible colors, and the bottom middle can only be <math>3</math> colors since it is next to the top left and bottom left. Similarly, we have <math>3</math> choices for the top right and <math>3</math> choices for the bottom right, which gives us a total of <math>5\cdot4\cdot3\cdot3\cdot3=\boxed{540}</math>.
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The top left rectangle can be <math>5</math> possible colors. Then the bottom left region can only be <math>4</math> possible colors, and the bottom middle can only be <math>3</math> colors since it is next to the top left and bottom left. Similarly, we have <math>3</math> choices for the top right and <math>3</math> choices for the bottom right, which gives us a total of <math>5\cdot4\cdot3\cdot3\cdot3=\boxed{\textbf{(D) }540}</math>.
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~Txu
 
~Txu
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==Solution 2 (Casework)==
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Case 1: All the rectangles are different colors. It would be <math>5! = 120</math> choices.
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Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us <math>5\cdot4\cdot3\cdot2 = 120</math>. But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of <math>360</math>.
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Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us <math>5\cdot4\cdot3 = 60</math>.
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Therefore, we have <math>120 + 360 + 60 = \boxed{\textbf{(D) }540}</math>.
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~[[OrenSH|orenbad]]
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==Video Solution 1 (Quick and Simple)==
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https://youtu.be/d_yfk7pYsSw
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~Education, the Study of Everything
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 +
==Video Solution 2 (Under a minute)==
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https://youtu.be/7yAh4MtJ8a8?si=88gRS7RBVbN7yCng&t=1290
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~Math-X
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==Video Solution 3==
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https://youtu.be/SRV4e74qsM8
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== See Also ==
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{{AMC10 box|year=2022|ab=A|num-b=8|num-a=10}}
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{{AMC12 box|year=2022|ab=A|num-b=6|num-a=8}}
 +
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 11:14, 25 December 2023

The following problem is from both the 2022 AMC 10A #9 and 2022 AMC 12A #7, so both problems redirect to this page.

Problem

A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?

[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]

$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$

Solution 1

The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{\textbf{(D) }540}$.

~Txu

Solution 2 (Casework)

Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.

Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$. But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$.

Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\cdot4\cdot3 = 60$.

Therefore, we have $120 + 360 + 60 = \boxed{\textbf{(D) }540}$.

~orenbad

Video Solution 1 (Quick and Simple)

https://youtu.be/d_yfk7pYsSw

~Education, the Study of Everything

Video Solution 2 (Under a minute)

https://youtu.be/7yAh4MtJ8a8?si=88gRS7RBVbN7yCng&t=1290

~Math-X

Video Solution 3

https://youtu.be/SRV4e74qsM8

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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