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

(Solution 4 (Casework on the Red Chips' Arrangements))
m (Solution 4 (Casework and Symmetry))
 
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<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math>
 
<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math>
  
==Solution 1==
+
==Solution 1 (Casework on the Center's Color Chip's Configurations)==
 
Call the different colors A,B,C. There are <math>3!=6</math> ways to rearrange these colors to these three letters, so <math>6</math> must be multiplied after the letters are permuted in the grid.  
 
Call the different colors A,B,C. There are <math>3!=6</math> ways to rearrange these colors to these three letters, so <math>6</math> must be multiplied after the letters are permuted in the grid.  
 
WLOG assume that A is in the center.  
 
WLOG assume that A is in the center.  
Line 37: Line 37:
 
  B & C & B     
 
  B & C & B     
 
\end{tabular}</cmath>
 
\end{tabular}</cmath>
This means that there are <math>4+2=6</math> 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  
+
This means that there are <math>4+2=6</math> ways to arrange A,B, and C in the grid, and there are <math>6</math> ways to rearrange the colors. Therefore, there are <math>6\cdot6=36</math> ways in total, which is  
<math>\boxed{\text{E}}</math>.
+
<math>\boxed{\textbf{(E)} ~36}</math>.
  
 
-happykeeper
 
-happykeeper
Line 47: Line 47:
 
<u><b>Case (1): The top-center and center-left chips have different colors.</b></u>
 
<u><b>Case (1): The top-center and center-left chips have different colors.</b></u>
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
unitsize(7mm);
 
unitsize(7mm);
 
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
 
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
Line 58: Line 59:
 
draw((8,0)--(8,3), linewidth(1.5));
 
draw((8,0)--(8,3), linewidth(1.5));
 
</asy>
 
</asy>
There are three sub-cases for Case (1):
+
There are three subcases for Case (1):
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
unitsize(7mm);
 
unitsize(7mm);
 
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
 
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
Line 109: Line 111:
 
draw((14,0)--(14,3), linewidth(1.5));
 
draw((14,0)--(14,3), linewidth(1.5));
 
</asy>
 
</asy>
As there are <math>3!=6</math> permutations of the three colors, each sub-case has <math>6</math> ways. So, Case (1) has <math>3\cdot6=18</math> ways in total.
+
As there are <math>3!=6</math> permutations of the three colors, each subcase has <math>6</math> ways. So, Case (1) has <math>3\cdot6=18</math> ways in total.
  
 
<u><b>Case (2): The top-center and center-left chips have the same color.</b></u>
 
<u><b>Case (2): The top-center and center-left chips have the same color.</b></u>
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
unitsize(7mm);
 
unitsize(7mm);
 
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
 
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
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draw((8,0)--(8,3), linewidth(1.5));
 
draw((8,0)--(8,3), linewidth(1.5));
 
</asy>
 
</asy>
There are three sub-cases for Case (2):
+
There are three subcases for Case (2):
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
unitsize(7mm);
 
unitsize(7mm);
 
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
 
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
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draw((14,0)--(14,3), linewidth(1.5));
 
draw((14,0)--(14,3), linewidth(1.5));
 
</asy>
 
</asy>
As there are <math>3!=6</math> permutations of the three colors, each sub-case has <math>6</math> ways. So, Case (2) has <math>3\cdot6=18</math> ways in total.
+
As there are <math>3!=6</math> permutations of the three colors, each subcase has <math>6</math> ways. So, Case (2) has <math>3\cdot6=18</math> ways in total.
  
 
<u><b>Answer</b></u>
 
<u><b>Answer</b></u>
Line 182: Line 186:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Solution 3 (Casework and Derangements)==
+
== Solution 3 (Casework on the Red Chips' Configurations) ==
<b>Case (1)</b>: We have a permutation of R, B, and G as all of the rows. There are <math>3!</math> ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, <math>\frac{3!}{e} \approx 2</math>, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are <math>2</math> possible permutations for the last row. Thus, there are <math>3!\cdot2\cdot2=24</math> possibilities.
+
We consider all possible configurations of the red chips for which rotations matter:
 
 
 
 
<b>Case (2)</b>: All of the rows have two chips that are the same color and one that is different. There are obviously <math>3</math> possible configurations for the first row, <math>2</math> for the second, and <math>2</math> for the third. <math>3\cdot2\cdot2=12</math>.
 
 
 
Therefore, our answer is <math>24+12=\boxed{\textbf{(E)} ~36}.</math>
 
 
 
~michaelchang1
 
 
 
== Solution 4 (Casework on the Red Chips' Configurations) ==
 
We will consider all possible configurations of the red chips, such that rotations create different configurations:
 
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
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draw((26,0)--(26,3), linewidth(1.5));
 
draw((26,0)--(26,3), linewidth(1.5));
  
label("$2$ rotations",(1.5,-0.75));
+
label("Rotational",(1.5,4.5));
label("$4$ rotations",(7.5,-0.75));
+
label("Symmetry",(1.5,3.75));
label("$4$ rotations",(13.5,-0.75));
+
label("$2$ Configurations",(1.5,-0.75));
label("$4$ rotations",(19.5,-0.75));
+
label("$4$ Configurations",(7.5,-0.75));
label("$4$ rotations",(25.5,-0.75));
+
label("$4$ Configurations",(13.5,-0.75));
 +
label("$4$ Configurations",(19.5,-0.75));
 +
label("$4$ Configurations",(25.5,-0.75));
 
</asy>
 
</asy>
As there are <math>2!=2</math> permutations of blue and green, the answer is <math>2\cdot(2+4+4+4+4)=\boxed{\textbf{(E)} ~36}.</math>  
+
As there are <math>2!=2</math> permutations of blue and green for each configuration, the answer is <math>2\cdot(2+4+4+4+4)=\boxed{\textbf{(E)} ~36}.</math>  
  
~MRENTHUSIASM (credit given to Flame Kho)
+
~MRENTHUSIASM (credit given to FlameKhoEmberish)
 +
 
 +
==Solution 4 (Casework and Symmetry)==
 +
 
 +
<math>(1) \quad</math><math>\begin{tabular}{ c c c }
 +
R & G & ? \\
 +
B & R & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
R & G & B \\
 +
B & R & G \\ 
 +
R & G & B   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
R & G & R \\
 +
B & R & B \\ 
 +
G & B & G   
 +
\end{tabular}</math>
 +
<math>\quad 3 \cdot 2 \cdot 2 = 12</math>
 +
 
 +
There are <math>3</math> choices for <math>R</math>, <math>2</math> choices for <math>G</math>. <math>R</math> on the down left corner can be switched with <math>B</math> on the upper right corner.
 +
 
 +
<math>(2) \quad</math><math>\begin{tabular}{ c c c }
 +
R & G & ? \\
 +
B & R & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
R & G & B \\
 +
B & R & G \\ 
 +
G & B & R   
 +
\end{tabular}</math>
 +
<math>\quad 3 \cdot 2 = 6</math>
 +
 
 +
There are <math>3</math> choices for <math>R</math>, <math>2</math> choices for <math>G</math>.
 +
 
 +
<math>(3) \quad</math><math>\begin{tabular}{ c c c }
 +
G & R & ? \\
 +
R & B & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
G & R & B \\
 +
R & B & G \\ 
 +
G & R & B   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
G & R & G \\
 +
R & B & R \\ 
 +
B & G & B   
 +
\end{tabular}</math>
 +
<math>\quad 3 \cdot 2 \cdot 2 = 12</math>
 +
 
 +
Note that <math>(3)</math> is a <math>180</math>° rotation of <math>(1)</math>.
 +
 
 +
<math>(4) \quad</math><math>\begin{tabular}{ c c c }
 +
G & R & ? \\
 +
R & B & ? \\ 
 +
? & ? & ?   
 +
\end{tabular}</math>
 +
<math>\Longrightarrow</math>
 +
<math>\begin{tabular}{ c c c }
 +
G & R & B \\
 +
R & B & G \\ 
 +
B & G & R   
 +
\end{tabular}</math>
 +
<math>\quad 3 \cdot 2 = 6</math>
 +
 
 +
Note that <math>(4)</math> is a <math>90</math>° rotation of <math>(2)</math>.
 +
 
 +
Therefore, the answer is <math>2 \cdot (12 + 6) = \boxed{\textbf{(E)} ~36}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 5 (Casework and Derangements)==
 +
<b>Case (1)</b>: We have a permutation of R, B, and G as all of the rows. There are <math>3!</math> ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, <math>\frac{3!}{e} \approx 2</math>, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are <math>2</math> possible permutations for the last row. Thus, there are <math>3!\cdot2\cdot2=24</math> possibilities.
 +
 
 +
<b>Case (2)</b>: All of the rows have two chips that are the same color and one that is different. There are obviously <math>3</math> possible configurations for the first row, <math>2</math> for the second, and <math>2</math> for the third. Thus, there are <math>3\cdot2\cdot2=12</math> possibilities.
 +
 
 +
Therefore, our answer is <math>24+12=\boxed{\textbf{(E)} ~36}.</math>
 +
 
 +
~michaelchang1
  
 
== Video Solution (Easiest) ==
 
== Video Solution (Easiest) ==
Line 262: Line 345:
 
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) ==
 
== Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations) ==
 
https://youtu.be/wKJ9ppI-8Ew ~ pi_is_3.14
 
https://youtu.be/wKJ9ppI-8Ew ~ pi_is_3.14
 +
== Video Solution by The Power of Logic==
 +
https://www.youtube.com/watch?v=TEsHuvXA9Ic
 +
 +
== Video Solution by MRENTHUSIASM (English & Chinese) ==
 +
https://www.youtube.com/watch?v=_2hCBZHb3SA
 +
 +
~MRENTHUSIASM
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2021|ab=A|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:37, 28 December 2021

Problem

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 1 (Casework on the Center's Color Chip's Configurations)

Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. WLOG assume that A is in the center. \[\begin{tabular}{ c c c }  ? & ? & ? \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In this configuration, there are two cases, either all the A's lie on the same diagonal: \[\begin{tabular}{ c c c }  ? & ? & A \\   ? & A & ? \\    A & ? & ?     \end{tabular}\] or all the other two A's are on adjacent corners: \[\begin{tabular}{ c c c }  A & ? & A \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.

In each case there is only one way to put the three B's and the three C's as shown in the diagrams. \[\begin{tabular}{ c c c }   C & B & A \\    B & A & C \\     A & C & B     \end{tabular}\] \[\begin{tabular}{ c c c }  A & B & A \\   C & A & C \\    B & C & B     \end{tabular}\] 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 $\boxed{\textbf{(E)} ~36}$.

-happykeeper

Solution 2 (Casework on the Top-Center and Center-Left Chips)

Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:

Case (1): The top-center and center-left chips have different colors. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); [/asy] There are three subcases for Case (1): [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, green); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green);  fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, blue); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue);  fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, green); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, blue); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, red); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, blue); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, green);  draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5));  draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));  draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); [/asy] As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (1) has $3\cdot6=18$ ways in total.

Case (2): The top-center and center-left chips have the same color. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); [/asy] There are three subcases for Case (2): [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, green); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red);  fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, green); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, red); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue);  fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, blue); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, green); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, red); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, blue); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, green); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red);  draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5));  draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));  draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); [/asy] As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (2) has $3\cdot6=18$ ways in total.

Answer

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

~MRENTHUSIASM

Solution 3 (Casework on the Red Chips' Configurations)

We consider all possible configurations of the red chips for which rotations matter: [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red);  fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, red); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red);  fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, red); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red);  fill((18,2)--(19,2)--(19,3)--(18,3)--cycle, red); fill((20,1)--(21,1)--(21,2)--(20,2)--cycle, red); fill((19,0)--(20,0)--(20,1)--(19,1)--cycle, red);  fill((24,1)--(25,1)--(25,2)--(24,2)--cycle, red); fill((26,1)--(27,1)--(27,2)--(26,2)--cycle, red); fill((25,2)--(26,2)--(26,3)--(25,3)--cycle, red);  draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5));  draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));  draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5));  draw((18,0)--(21,0)--(21,3)--(18,3)--cycle, linewidth(1.5)); draw((18,1)--(21,1), linewidth(1.5)); draw((18,2)--(21,2), linewidth(1.5)); draw((19,0)--(19,3), linewidth(1.5)); draw((20,0)--(20,3), linewidth(1.5));  draw((24,0)--(27,0)--(27,3)--(24,3)--cycle, linewidth(1.5)); draw((24,1)--(27,1), linewidth(1.5)); draw((24,2)--(27,2), linewidth(1.5)); draw((25,0)--(25,3), linewidth(1.5)); draw((26,0)--(26,3), linewidth(1.5));  label("Rotational",(1.5,4.5)); label("Symmetry",(1.5,3.75)); label("$2$ Configurations",(1.5,-0.75)); label("$4$ Configurations",(7.5,-0.75)); label("$4$ Configurations",(13.5,-0.75)); label("$4$ Configurations",(19.5,-0.75)); label("$4$ Configurations",(25.5,-0.75)); [/asy] As there are $2!=2$ permutations of blue and green for each configuration, the answer is $2\cdot(2+4+4+4+4)=\boxed{\textbf{(E)} ~36}.$

~MRENTHUSIASM (credit given to FlameKhoEmberish)

Solution 4 (Casework and Symmetry)

$(1) \quad$$\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    R & G & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & R \\   B & R & B \\    G & B & G     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$

There are $3$ choices for $R$, $2$ choices for $G$. $R$ on the down left corner can be switched with $B$ on the upper right corner.

$(2) \quad$$\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    G & B & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$

There are $3$ choices for $R$, $2$ choices for $G$.

$(3) \quad$$\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    G & R & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & G \\   R & B & R \\    B & G & B     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$

Note that $(3)$ is a $180$° rotation of $(1)$.

$(4) \quad$$\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    B & G & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$

Note that $(4)$ is a $90$° rotation of $(2)$.

Therefore, the answer is $2 \cdot (12 + 6) = \boxed{\textbf{(E)} ~36}$.

~isabelchen

Solution 5 (Casework and Derangements)

Case (1): We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$ possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.

Case (2): All of the rows have two chips that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. Thus, there are $3\cdot2\cdot2=12$ possibilities.

Therefore, our answer is $24+12=\boxed{\textbf{(E)} ~36}.$

~michaelchang1

Video Solution (Easiest)

https://www.youtube.com/watch?v=UPUrYN1YuVA ~ MathEx

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

https://youtu.be/wKJ9ppI-8Ew ~ pi_is_3.14

Video Solution by The Power of Logic

https://www.youtube.com/watch?v=TEsHuvXA9Ic

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=_2hCBZHb3SA

~MRENTHUSIASM

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
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Problem 24
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