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Difference between revisions of "2021 AMC 10A Problems/Problem 25"

(→‎Solution)
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-happykeeper
 
-happykeeper
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==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.
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & ? \\
 +
G & R & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</cmath>
 +
Each placement has 6 permutations, as there are <math>3!=6</math> ways to permute RBG.
 +
 +
There are three sub-cases for Case (1):
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & R \\
 +
G & R & G \\ 
 +
B & G & B   
 +
\end{tabular}</cmath>
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & G \\
 +
G & R & B \\ 
 +
R & B & G   
 +
\end{tabular}</cmath>
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & G \\
 +
G & R & B \\ 
 +
R & G & R   
 +
\end{tabular}</cmath>
 +
So, Case (1) has <math>3\cdot6=18</math> ways.
 +
 +
 +
(2) The two balls adjacent to the top-left red ball have the same color.
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & ? \\
 +
B & ? & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</cmath>
 +
Each placement has 6 permutations, as there are <math>\binom32\binom21=6</math> ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB).
 +
There are three sub-cases for Case (2):
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & R \\
 +
B & G & B \\ 
 +
G & R & G   
 +
\end{tabular}</cmath>
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & G \\
 +
B & G & R \\ 
 +
R & B & G   
 +
\end{tabular}</cmath>
 +
<cmath>\begin{tabular}{ c c c }
 +
R & B & G \\
 +
B & G & R \\ 
 +
G & R & B   
 +
\end{tabular}</cmath>
 +
So, Case (2) has <math>3\cdot6=18</math> ways.
 +
 +
 +
Together, the answer is <math>18+18=\boxed{\textbf{(E)} ~36}.</math>
 +
 +
~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) ==
 
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) ==

Revision as of 03:34, 12 February 2021

Problem 25

How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally.

$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution

Call the different colors A,B,C. There are $3!=6$ 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 $6\cdot6=36$ ways in total, which is $\boxed{\text{E}}$.

(someone who knows asy please help me insert diagrams, thanks in advance!)

-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. \[\begin{tabular}{ c c c } R & B & ? \\ G & R & ? \\  ? & ? & ? \end{tabular}\] Each placement has 6 permutations, as there are $3!=6$ ways to permute RBG.

There are three sub-cases for Case (1): \[\begin{tabular}{ c c c } R & B & R \\ G & R & G \\ B & G & B \end{tabular}\] \[\begin{tabular}{ c c c } R & B & G \\ G & R & B \\ R & B & G \end{tabular}\] \[\begin{tabular}{ c c c } R & B & G \\ G & R & B \\ R & G & R \end{tabular}\] So, Case (1) has $3\cdot6=18$ ways.


(2) The two balls adjacent to the top-left red ball have the same color. \[\begin{tabular}{ c c c } R & B & ? \\ B & ? & ? \\  ? & ? & ? \end{tabular}\] Each placement has 6 permutations, as there are $\binom32\binom21=6$ ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2): \[\begin{tabular}{ c c c } R & B & R \\ B & G & B \\ G & R & G \end{tabular}\] \[\begin{tabular}{ c c c } R & B & G \\ B & G & R \\ R & B & G \end{tabular}\] \[\begin{tabular}{ c c c } R & B & G \\ B & G & R \\ G & R & B \end{tabular}\] So, Case (2) has $3\cdot6=18$ ways.


Together, the answer is $18+18=\boxed{\textbf{(E)} ~36}.$

~MRENTHUSIASM

Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)

https://youtu.be/wKJ9ppI-8Ew

~ pi_is_3.14

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