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Difference between revisions of "2023 AMC 10A Problems/Problem 20"
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Each square in a <math>3 x 3</math> grid of sqaures is colored red, white, blue or green so every <math>2 x 2</math> sqare contains one sqaure of each color. One such coloring is shown below(help me with the picture). How many diffrent colorings are possible?
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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?  
  
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Solution 1.
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==Solution 1==
  
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WLOG, we start with the TL grid, we have 4 options. Then three adjacent grids have 3,2,1 options.
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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 (let a "tile" denote a <math>1\times1</math> square and "square" refer to <math>2\times2</math>). 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.  
  
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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.
  
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Tecnodoggo~
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~Technodoggo

Revision as of 20:23, 9 November 2023

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?

Solution 1

We first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile (let a "tile" denote a $1\times1$ square and "square" refer to $2\times2$). 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 $3\cdot4!=72$ ways.

~Technodoggo

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