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

Revision as of 23:06, 12 November 2023

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

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

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

Let us start with choosing the colors of the top middle square and the center square. There is $4$ ways to choose the color of the top middle square, and $3$ 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 $2$ ways to choose the color, then there are also $2$ ways to choose the color of the bottom middle square, because it has two things it cannot be.

Case 2: is that the middle left and middle right squares are different colors. There are $2!$ ways to order the two different colors, and then there is $1$ way to choose the two different colors.

Thus, once we choose the middle squares (in 12 possible ways), we have 6 ways to color the squares. Thus, we have $6 \cdot 12=72$ , and so the answer is $48+24=\boxed{\textbf{(D) }72}$ .

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: $4(6+12)=\boxed{\textbf{(D) }72}$ .

Video Solution 1

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

-paixiao

Video Solution by OmegaLearn

https://youtu.be/Zrqy5yGYlvQ

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. $(D)$

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