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Difference between revisions of "2023 AMC 10A Problems/Problem 20"

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==Problem==
 +
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?
  
 +
<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>
 +
 +
==Solution 1==
 +
 +
Let a "tile" denote a \(1 \times 1\) square, and a "square" refer to a \(2 \times 2\) square.
 +
 +
We have \(4! = 24\) possible ways to fill out the top-left square. Next, we fill out the bottom-right tile. In the bottom-right square, one corner is already filled (from our initial coloring), so we have 3 options remaining.
 +
 +
For the right-middle tile, it is part of two squares: the top-right and the top-left. Among these squares, 3 colors have already been used, leaving us with only 1 remaining option. Similarly, every other square has only one available option for coloring.
 +
 +
Thus, the total number of ways is \(3 \times 4! = \boxed{\textbf{(D) }72}\).
 +
 +
~ stevehan
 +
 +
~ 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 <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==
 +
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.
 +
Then, we can choose <math>1</math> color to be in the center. <math>{4 \choose 1}</math> = 4
 +
 +
Now, we have some casework:
 +
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges.
 +
<math>232</math>
 +
<math>414</math>
 +
<math>232</math>
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There's <math>3!=6</math> 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.
 +
<math>232</math>
 +
<math>414</math>
 +
<math>323</math>
 +
The orientation of the 2 colors has 2 possibilities, and there are <math>3!</math> color permutations. <math>2*3!=12</math>
 +
 +
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 <math>2*2</math> square.
 +
We multiply the start with the sum of the 2 cases: <math>4(6+12)=\boxed{\textbf{(D) }72}</math>.
 +
 +
 +
==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
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=YXUPq1RZv92E0d8H&t=7113
 +
 +
~Math-X
 +
 +
==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 <math>4\cdot 3\cdot 2\cdot 2=48</math> 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
 +
<math>4\cdot 3\cdot 2=24</math> ways.
 +
 +
The answer= 48+24=72. <math>(D)</math>
 +
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=A|num-b=19|num-a=21}}
 +
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 23:09, 30 September 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 \(1 \times 1\) square, and a "square" refer to a \(2 \times 2\) square.

We have \(4! = 24\) possible ways to fill out the top-left square. Next, we fill out the bottom-right tile. In the bottom-right square, one corner is already filled (from our initial coloring), so we have 3 options remaining.

For the right-middle tile, it is part of two squares: the top-right and the top-left. Among these squares, 3 colors have already been used, leaving us with only 1 remaining option. Similarly, every other square has only one available option for coloring.

Thus, the total number of ways is \(3 \times 4! = \boxed{\textbf{(D) }72}\).

~ stevehan

~ 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 Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=YXUPq1RZv92E0d8H&t=7113

~Math-X

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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