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

(Solution 1)
(Solution 2)
Line 17: Line 17:
 
Note that there can be no overlap between colors in each square.
 
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\c1=4</math>
 
Then, we can choose <math>1</math> color to be in the center. <math>4\c1=4</math>
 +
 
Now, we have some casework:
 
Now, we have some casework:
 
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges.
 
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges.
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<math>232</math>
 
<math>232</math>
 
There's <math>3!=6</math> ways to do this.
 
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.
 
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>232</math>

Revision as of 20:32, 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?

2023 10a 20.png

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 $3\cdot4!=72$ ways.

~Technodoggo

Solution 2

Note that there can be no overlap between colors in each square. Then, we can choose $1$ color to be in the center. $4\c1=4$

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