Difference between revisions of "2023 AMC 10A Problems/Problem 20"
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We first have <math>4!=24</math> 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 <math>3</math> options left to pick from. | We first have <math>4!=24</math> 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 <math>3</math> 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, <math>3</math> 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 <math>3\cdot4!=72</math> ways. | + | We then look at the right middle tile. It is part of two squares: the top right and top left. Among these squares, <math>3</math> 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 <math>3\cdot4!=\boxed{\textbf{(D) }72}</math> ways. |
~Technodoggo | ~Technodoggo | ||
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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. | 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{72}</math>. | + | We multiply the start with the sum of the 2 cases: <math>4(6+12)=\boxed{\textbf{(D) }72}</math>. |
==See Also== | ==See Also== | ||
{{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}} | ||
Revision as of 21:24, 9 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
Note that there can be no overlap between colors in each square.
Then, we can choose 
color to be in the center. 
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: 
.
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. 
