Difference between revisions of "2023 AMC 10A Problems/Problem 20"
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==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> | ||
| + | ~SirAppel | ||
| + | ==Solution 3== | ||
Note that there can be no overlap between colors in each square. | Note that there can be no overlap between colors in each square. | ||
| Line 37: | Line 44: | ||
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 | + | ==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. | 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 | Case 1 : C=Red and F=Blue | ||
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-paixiao | -paixiao | ||
| − | ==Solution | + | ==Solution 5== |
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: | 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: | ||
Revision as of 17:54, 12 November 2023
Contents
Problem
Each square in a 
grid of squares is colored red, white, blue, or green so that every 
square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
Solution 1
Let a "tile" denote a 
square and "square" refer to .
We first have 
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 
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, 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 
ways.
~Technodoggo
Solution 2
First, we must choose the color of the center square. We have 
ways to do this.
Then, we have 
ways to order the top left squares to make them all unique.
Finally, we have ways to order the bottom right 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 
~SirAppel
Solution 3
Note that there can be no overlap between colors in each square.
Then, we can choose 
color to be in the center. 
= 4
Now, we have some casework:
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges.
There's 
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.
The orientation of the 2 colors has 2 possibilities, and there are 
color permutations. 
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 
square.
We multiply the start with the sum of the 2 cases: 
.
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. 
.
-paixiao
Solution 5
Let us start with choosing the colors of the top middle square and the center square. There is ways to choose the color of the top middle square, and 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 
ways to choose the color, then there are also 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 
ways to order the two different colors, and then there is 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 
, and so the answer is 
.
Video Solution 1
https://www.youtube.com/watch?v=VWZBpT9lt0Q&t=6s
-paixiao
Video Solution by OmegaLearn
Video Solution by Solve It (Simple)
https://www.youtube.com/watch?v=ke4ZOV4KA6Y
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~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 
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
ways.
The answer= 48+24=72. 
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. 
