2021 AMC 10A Problems/Problem 25
Contents
Problem
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
.
-happykeeper
Solution 2 (Casework)
Without the loss of generality, we place a red ball in the top-left square. There are two cases:
(1) The two balls adjacent to the top-left red ball have different colors.
![\[\begin{tabular}{ c c c } R & B & ? \\ G & R & ? \\ ? & ? & ? \end{tabular}\]](http://latex.artofproblemsolving.com/c/6/f/c6f6e68d3d0a4649d721f88df433328612a62a3c.png)
Each placement has 6 permutations, as there are ways to permute RBG.
There are three sub-cases for Case (1):
![\[\begin{tabular}{ c c c } R & B & R \\ G & R & G \\ B & G & B \end{tabular}\]](http://latex.artofproblemsolving.com/4/3/d/43d1ef0f581ac20e950c42de42db8fbbd8980ce1.png)
![\[\begin{tabular}{ c c c } R & B & G \\ G & R & B \\ R & B & G \end{tabular}\]](http://latex.artofproblemsolving.com/1/9/6/196128bb02afbf506ae90c4c5dfb1e12d9f92f88.png)
![\[\begin{tabular}{ c c c } R & B & G \\ G & R & B \\ R & G & R \end{tabular}\]](http://latex.artofproblemsolving.com/f/1/9/f19a41cdca3f848c0d18ef16e609fd9104e89a25.png)
So, Case (1) has 
ways.
(2) The two balls adjacent to the top-left red ball have the same color.
![\[\begin{tabular}{ c c c } R & B & ? \\ B & ? & ? \\ ? & ? & ? \end{tabular}\]](http://latex.artofproblemsolving.com/a/5/f/a5f9ec83fd2d6e541d2db9fea6297257167fa04b.png)
Each placement has 6 permutations, as there are 
ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB).
There are three sub-cases for Case (2):
![\[\begin{tabular}{ c c c } R & B & R \\ B & G & B \\ G & R & G \end{tabular}\]](http://latex.artofproblemsolving.com/7/9/3/793748f2e2d6de4f8c6d799aa304e1e59698106e.png)
![\[\begin{tabular}{ c c c } R & B & G \\ B & G & R \\ R & B & G \end{tabular}\]](http://latex.artofproblemsolving.com/1/4/f/14f795fd31753347f221daf3f05f798e670c9adc.png)
![\[\begin{tabular}{ c c c } R & B & G \\ B & G & R \\ G & R & B \end{tabular}\]](http://latex.artofproblemsolving.com/5/d/9/5d9c7fc4979b5e945a688244283d29951b41a2bf.png)
So, Case (2) has ways.
Together, the answer is 
~MRENTHUSIASM
Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
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| All AMC 10 Problems and Solutions | ||
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