Difference between revisions of "2021 AMC 10A Problems/Problem 25"
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| + | ==Problem 25== | ||
| + | How many ways are there to place <math>3</math> indistinguishable red chips, <math>3</math> indistinguishable blue chips, and <math>3</math> indistinguishable green chips in the squares of a <math>3 \times 3</math> grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally. | ||
| + | |||
| + | <math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math> | ||
| + | ==Solution== | ||
| + | Call the different colors A,B,C. There are <math>3!=6</math> ways to rearrange these colors to these three letters, so we must remember to multiply by 6. | ||
| + | We can assume that A is in the center. In this configuration, the other 2 A's must either lie on the same diagonal or lie in adjacent corner squares. There are two ways to place the A's in the former, while there are four ways to place the A's in the latter. In each case there is only one way to put the three B's and the three C's as shown in the diagrams. This means that there are 4+2=6 ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are <math>6\cdot6=36</math> ways in total, which is | ||
| + | <math>\boxed{\text{E}}</math> | ||
| + | (someone who knows asy please help me insert diagrams, thanks in advance!) | ||
| + | -happykeeper | ||
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) == | == Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) == | ||
https://youtu.be/wKJ9ppI-8Ew | https://youtu.be/wKJ9ppI-8Ew | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
Revision as of 00:48, 12 February 2021
Problem 25
How many ways are there to place 
indistinguishable red chips, indistinguishable blue chips, and indistinguishable green chips in the squares of a 
grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.
Solution
Call the different colors A,B,C. There are 
ways to rearrange these colors to these three letters, so we must remember to multiply by 6.
We can assume that A is in the center. In this configuration, the other 2 A's must either lie on the same diagonal or lie in adjacent corner squares. There are two ways to place the A's in the former, while there are four ways to place the A's in the latter. In each case there is only one way to put the three B's and the three C's as shown in the diagrams. This means that there are 4+2=6 ways to arrange A,B, and C in the grid, and there are 6 ways to rearrange the colors. Therefore, there are 
ways in total, which is
(someone who knows asy please help me insert diagrams, thanks in advance!)
-happykeeper
Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
~ pi_is_3.14
