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Difference between revisions of "2023 AMC 10A Problems/Problem 20"
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Each square in a <math>3\times3</math> grid of squares is colored red, white, blue, or green so that every <math>2\times2</math> square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?  
 
Each square in a <math>3\times3</math> grid of squares is colored red, white, blue, or green so that every <math>2\times2</math> square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?  
  
−
[[Image:2023_10a_20.png]]
+
<asy>
 +
unitsize(0.5cm, 0.5cm);
 +
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
 +
draw((0,3)--(9,3));
 +
draw((0,6)--(9,6));
 +
draw((3,0)--(3,9));
 +
draw((6,0)--(6,9));
 +
 
 +
draw((18,0)--(27,0)--(27,9)--(18,9)--cycle);
 +
draw((18,3)--(27,3));
 +
draw((18,6)--(27,6));
 +
draw((21,0)--(21,9));
 +
draw((24,0)--(24,9));
 +
 
 +
label("R", (19.5,1.5));
 +
label("B", (22.5,1.5));
 +
label("R", (25.5,1.5));
 +
 
 +
label("G", (19.5,4.5));
 +
label("W", (22.5,4.5));
 +
label("G", (25.5,4.5));
 +
 
 +
label("B", (19.5,7.5));
 +
label("R", (22.5,7.5));
 +
label("B", (25.5,7.5));
 +
 
 +
</asy>
  
 
<math>\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96</math>
 
<math>\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96</math>
Line 17: Line 43:
  
 
==Solution 2==
 
==Solution 2==
−
First, we must choose the color of the center square. We have <math>4</math> ways to do this.
 
−
Then, we have <math>6</math> ways to order the top left <math>3</math> squares to make them all unique.
 
−
Finally, we have <math>3</math> ways to order the bottom right <math>3</math> squares, such that their order doesn't collide with the top left square to make the top right and bottom left squares non-distinct.
 
  
−
Hence, our answer is <math>4 \cdot 6 \cdot 3=\boxed{\textbf{(D) }72}</math>
+
<asy>
−
~SirAppel
+
unitsize(0.5cm, 0.5cm);
 +
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
 +
draw((0,3)--(9,3));
 +
draw((0,6)--(9,6));
 +
draw((3,0)--(3,9));
 +
draw((6,0)--(6,9));
 +
 
 +
label("R", (1.5,1.5));
 +
label("B", (4.5,1.5));
 +
label("R", (7.5,1.5));
 +
 
 +
label("G", (1.5,4.5));
 +
label("W", (4.5,4.5));
 +
label("G", (7.5,4.5));
 +
 
 +
label("B", (1.5,7.5));
 +
label("R", (4.5,7.5));
 +
label("B", (7.5,7.5));
 +
 
 +
 
 +
 
 +
draw((18,0)--(27,0)--(27,9)--(18,9)--cycle);
 +
draw((18,3)--(27,3));
 +
draw((18,6)--(27,6));
 +
draw((21,0)--(21,9));
 +
draw((24,0)--(24,9));
 +
 
 +
label("R", (19.5,1.5));
 +
label("B", (22.5,1.5));
 +
label("R", (25.5,1.5));
 +
 
 +
label("G", (19.5,4.5));
 +
label("W", (22.5,4.5));
 +
label("G", (25.5,4.5));
 +
 
 +
label("R", (19.5,7.5));
 +
label("B", (22.5,7.5));
 +
label("R", (25.5,7.5));
 +
 
 +
</asy>
 +
We can split this problem into <math>2</math> cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.
 +
 
 +
Square 1:
 +
The first square can be rotated to create another square so we have to multiply the number of arrangements by <math>2</math>. We have <math>4! = 24</math> arrangements without rotating and <math>24\cdot 2 = 48</math> arrangements in total for the first square.
 +
 
 +
Square 2:
 +
There are <math>4! = 24</math> ways to arrange the colors.
 +
 
 +
In total, we have <math>48 + 24 = \boxed{\textbf{(D) }72}</math> arrangements.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:South South] (LaTeX and Solution)
 +
Edit by: Mismatchedcubing/Andrew_Lu
 +
 
 
==Solution 3==
 
==Solution 3==
 +
Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.
 +
Case 1 : C=Red and F=Blue
 +
In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways.
 +
Case 2 : C=Blue and F=Red
 +
In this case, one of G and H needs to be White and Red, and  H and I needs to be White and Blue. There is 1 way to color this.
 +
In total, we get 24*(2+1)=72 ways to color the grid. <math>\boxed{\textbf{(D) }72}</math>.
  
 +
-paixiao
 +
 +
==Solution 4==
 +
We will choose colors step-by-step:
 +
 +
1. There are <math>4</math> ways to choose a color in the center.
 +
 +
2. Then we select any corner and there would be <math>3</math> ways to choose a color as we can't use the same color as the one in the center.
 +
 +
3. Consider the <math>2\times 2</math> square that contains the center and the corner we have selected. For the other <math>2</math> squares, there are <math>2</math> ways to choose colors.
 +
 +
4. Now, consider how many configurations it makes sense to construct the <math>2\times 2</math> square opposite to the corner we have selected using the <math>2</math> other <math>2\times 2</math> squares, and we get <math>3</math> configurations.
 +
 +
Finally, the answer is <math>4 \cdot 3 \cdot 2 \cdot 3 = \boxed{\textbf{(D) }72}</math>
 +
 +
~jjaamm
 +
 +
==Solution 5==
 
Note that there can be no overlap between colors in each square.
 
Note that there can be no overlap between colors in each square.
 
Then, we can choose <math>1</math> color to be in the center. <math>{4 \choose 1}</math> = 4
 
Then, we can choose <math>1</math> color to be in the center. <math>{4 \choose 1}</math> = 4
Line 44: Line 143:
 
We multiply the start with the sum of the 2 cases: <math>4(6+12)=\boxed{\textbf{(D) }72}</math>.
 
We multiply the start with the sum of the 2 cases: <math>4(6+12)=\boxed{\textbf{(D) }72}</math>.
  
−
==Solution 4==
 
−
Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.
 
−
Case 1 : C=Red and F=Blue
 
−
In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways.
 
−
Case 2 : C=Blue and F=Red
 
−
In this case, one of G and H needs to be White and Red, and  H and I needs to be White and Blue. There is 1 way to color this.
 
−
In total, we get 24*(2+1)=72 ways to color the grid. <math>\boxed{\textbf{(D) }72}</math>.
 
  
−
-paixiao
+
==Solution 6==
 +
 
 +
<asy>
 +
unitsize(0.5cm, 0.5cm);
 +
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
 +
draw((0,3)--(9,3));
 +
draw((0,6)--(9,6));
 +
draw((3,0)--(3,9));
 +
draw((6,0)--(6,9));
 +
 
 +
label("2", (1.5,1.5));
 +
label("1", (4.5,1.5));
 +
label("1", (7.5,1.5));
 +
 
 +
label("2", (1.5,4.5));
 +
label("3", (4.5,4.5));
 +
label("1", (7.5,4.5));
 +
 
 +
label("3", (1.5,7.5));
 +
label("1", (4.5,7.5));
 +
label("2", (7.5,7.5));
 +
 
 +
</asy>
 +
 
 +
 
 +
Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, <math>2\times2\times2\times3\time3\times3</math> = <math>\boxed{\textbf{(D) }72}</math> (Similar to Solution 1)
 +
 
 +
-MEZE_RUN
  
−
==Solution 5==
+
==Video Solution by MegaMath==
−
Let us start with choosing the colors of the top middle square and the center square. There is <math>4</math> ways to choose the color of the top middle square, and <math>3</math> ways to choose the color of the center square, since these two squares must have two different colors. Then, from here, we have two remaining colors for us to put in the top left and middle left, and the same two colors to put the top right and middle right squares. Now, we have two cases:
 
  
−
Case 1: The middle left and middle right squares are the same color. There are <math>2</math> ways to choose the color, then there are also <math>2</math> ways to choose the color of the bottom middle square, because it has two things it cannot be.
+
https://www.youtube.com/watch?v=AhqOj8502Dc&t=65s
  
−
Case 2: is that the middle left and middle right squares are different colors. There are <math>2!</math> ways to order the two different colors, and then there is <math>1</math> way to choose the two different colors.
+
~megahertz13
  
−
Thus, once we choose the middle squares (in 12 possible ways), we have 6 ways to color the squares. Thus, we have <math>6 \cdot 12=72</math>, and so the answer is <math>48+24=\boxed{\textbf{(D) }72}</math>.
+
==Video Solution by Power Solve (easy to digest!)==
 +
https://youtu.be/0_eRTysDEsI
  
 
==Video Solution 1==
 
==Video Solution 1==
Line 70: Line 189:
 
==Video Solution by OmegaLearn ==
 
==Video Solution by OmegaLearn ==
 
https://youtu.be/Zrqy5yGYlvQ
 
https://youtu.be/Zrqy5yGYlvQ
 +
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 +
https://www.youtube.com/watch?v=vlaZ5P8UZls
  
 
==Video Solution by Solve It (Simple) ==
 
==Video Solution by Solve It (Simple) ==
Line 97: Line 220:
 
{{AMC10 box|year=2023|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2023|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Revision as of 11:23, 2 August 2024

Problem

Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?

[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]

$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$

Solution 1

Let a "tile" denote a $1\times1$ square and "square" refer to $2\times2$.

We first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options left to pick from.

We then look at the right middle tile. It is part of two squares: the top right and top left. Among these squares, $3$ colors have already been used, so we only have one more option for it. Similarly, every other square only has one more option, so we have a total of $3\cdot4!=\boxed{\textbf{(D) }72}$ ways.

~Technodoggo

Solution 2

[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("R", (1.5,1.5)); label("B", (4.5,1.5)); label("R", (7.5,1.5)); label("G", (1.5,4.5)); label("W", (4.5,4.5)); label("G", (7.5,4.5)); label("B", (1.5,7.5)); label("R", (4.5,7.5)); label("B", (7.5,7.5)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("R", (19.5,7.5)); label("B", (22.5,7.5)); label("R", (25.5,7.5)); [/asy] We can split this problem into $2$ cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.

Square 1: The first square can be rotated to create another square so we have to multiply the number of arrangements by $2$. We have $4! = 24$ arrangements without rotating and $24\cdot 2 = 48$ arrangements in total for the first square.

Square 2: There are $4! = 24$ ways to arrange the colors.

In total, we have $48 + 24 = \boxed{\textbf{(D) }72}$ arrangements.

~South (LaTeX and Solution) Edit by: Mismatchedcubing/Andrew_Lu

Solution 3

Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors. Case 1 : C=Red and F=Blue In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways. Case 2 : C=Blue and F=Red In this case, one of G and H needs to be White and Red, and H and I needs to be White and Blue. There is 1 way to color this. In total, we get 24*(2+1)=72 ways to color the grid. $\boxed{\textbf{(D) }72}$.

-paixiao

Solution 4

We will choose colors step-by-step:

1. There are $4$ ways to choose a color in the center.

2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.

3. Consider the $2\times 2$ square that contains the center and the corner we have selected. For the other $2$ squares, there are $2$ ways to choose colors.

4. Now, consider how many configurations it makes sense to construct the $2\times 2$ square opposite to the corner we have selected using the $2$ other $2\times 2$ squares, and we get $3$ configurations.

Finally, the answer is $4 \cdot 3 \cdot 2 \cdot 3 = \boxed{\textbf{(D) }72}$

~jjaamm

Solution 5

Note that there can be no overlap between colors in each square. Then, we can choose $1$ color to be in the center. ${4 \choose 1}$ = 4

Now, we have some casework: Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges. $232$ $414$ $232$ There's $3!=6$ ways to do this.

Case 2: 2 colors are placed with 2 in adjacent corners and 1 edge opposite them. The final color is placed in the remaining 2 edges. $232$ $414$ $323$ The orientation of the 2 colors has 2 possibilities, and there are $3!$ color permutations. $2*3!=12$

There can't be any more ways to do this, as we have combined all cases such that each color is used once and only once per $2*2$ square. We multiply the start with the sum of the 2 cases: $4(6+12)=\boxed{\textbf{(D) }72}$.


Solution 6

[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("2", (1.5,1.5)); label("1", (4.5,1.5)); label("1", (7.5,1.5)); label("2", (1.5,4.5)); label("3", (4.5,4.5)); label("1", (7.5,4.5)); label("3", (1.5,7.5)); label("1", (4.5,7.5)); label("2", (7.5,7.5)); [/asy]


Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, $2\times2\times2\times3\time3\times3$ = $\boxed{\textbf{(D) }72}$ (Similar to Solution 1)

-MEZE_RUN

Video Solution by MegaMath

https://www.youtube.com/watch?v=AhqOj8502Dc&t=65s

~megahertz13

Video Solution by Power Solve (easy to digest!)

https://youtu.be/0_eRTysDEsI

Video Solution 1

https://www.youtube.com/watch?v=VWZBpT9lt0Q&t=6s

-paixiao

Video Solution by OmegaLearn

https://youtu.be/Zrqy5yGYlvQ

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=vlaZ5P8UZls

Video Solution by Solve It (Simple)

https://www.youtube.com/watch?v=ke4ZOV4KA6Y

Video Solution

https://youtu.be/8ki0D3m_mEo

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://youtu.be/BDNsMEIzHF0

~IceMatrix Solution 5. 

Let's name the cells A,B,C,D,E,F,G,H,I from the top left to the bottom right.

Case 1. Cell B and cell H have the same color. The middle one cell E has 4 choices, cell B has 3 choices, then cell E has 2 choices and cell F has 2 choices, this gives $4\cdot 3\cdot 2\cdot 2=48$ ways.

Case 2. Cell B and cell H have different colors. The middle one cell E has 4 choices, cell B has 3 choices, cell H has 2 choices, then cell D and F each can only have one choice(different from B,E,H). This gives $4\cdot 3\cdot 2=24$ ways.

The answer= 48+24=72. $(D)$

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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. AMC logo.png

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