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

(Undo revision 151169 by MRENTHUSIASM (talk))
(Tag: Undo)
m (Solution 4 (Casework on the Red Chips' Configurations))
 
(30 intermediate revisions by the same user not shown)
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-happykeeper
 
-happykeeper
  
==Solution 2 (Casework)==
+
==Solution 2 (Casework on the Top-Center and Center-Left Chips)==
Without the loss of generality, we place a red ball in the top-left square. There are two cases:
+
Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
  
<u><b>Case (1): The two balls adjacent to the top-left red ball have different colors.</b></u>
+
<u><b>Case (1): The top-center and center-left chips have different colors.</b></u>
<cmath>\begin{array}{|c|c|c|}
+
<asy>
\hline
+
/* Made by MRENTHUSIASM */
& & \hspace{3.25mm} \\ [-2.25ex]
+
unitsize(7mm);
\mathbf{R} & \mathbf{B} &  \\
+
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
\hline
+
fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue);
& & \\ [-2.25ex]
+
fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green);
\mathbf{G} & \mathbf{R} &  \\ 
+
fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red);
\hline
+
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
& & \\ [0.5ex] 
+
draw((6,1)--(9,1), linewidth(1.5));
\hline
+
draw((6,2)--(9,2), linewidth(1.5));
\end{array}</cmath>
+
draw((7,0)--(7,3), linewidth(1.5));
Each placement has <math>6</math> permutations, as there are <math>3!=6</math> ways to permute RBG.
+
draw((8,0)--(8,3), linewidth(1.5));
 +
</asy>
 +
There are three sub-cases 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));
  
There are three sub-cases for Case (1):
+
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
<cmath>\begin{array}{ccccccc}
+
draw((6,1)--(9,1), linewidth(1.5));
\begin{array}{|c|c|c|}
+
draw((6,2)--(9,2), linewidth(1.5));
\hline
+
draw((7,0)--(7,3), linewidth(1.5));
\hspace{3.4mm} & \hspace{3.4mm} & \hspace{3.4mm} \\ [-2.25ex]
+
draw((8,0)--(8,3), linewidth(1.5));
\mathrm{R} & \mathrm{B} & \mathrm{R} \\ 
+
 
\hline
+
draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5));
& & \\ [-2.25ex]
+
draw((12,1)--(15,1), linewidth(1.5));
\mathrm{G} & \mathrm{R} & \mathrm{G} \\     
+
draw((12,2)--(15,2), linewidth(1.5));
\hline
+
draw((13,0)--(13,3), linewidth(1.5));
& & \\ [-2.25ex]
+
draw((14,0)--(14,3), linewidth(1.5));
\mathrm{B} & \mathrm{G} & \mathrm{B}  \\
+
</asy>
\hline
+
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.
\end{array}
 
& & &
 
\begin{array}{|c|c|c|}
 
\hline
 
\hspace{3.4mm} & \hspace{3.4mm} & \hspace{3.4mm} \\ [-2.25ex]
 
\mathrm{R} & \mathrm{B} & \mathrm{G} \\ 
 
\hline
 
& & \\ [-2.25ex]
 
\mathrm{G} & \mathrm{R} & \mathrm{B} \\ 
 
\hline
 
& & \\ [-2.25ex]
 
\mathrm{R} & \mathrm{B} & \mathrm{G} \\
 
\hline
 
\end{array}
 
& & &
 
\begin{array}{|c|c|c|}
 
\hline
 
\hspace{3.4mm} & \hspace{3.4mm} & \hspace{3.4mm} \\ [-2.25ex]
 
\mathrm{R} & \mathrm{B} & \mathrm{G} \\ 
 
\hline
 
& & \\ [-2.25ex]
 
\mathrm{G} & \mathrm{R} & \mathrm{B} \\ 
 
\hline
 
& & \\ [-2.25ex]
 
\mathrm{B} & \mathrm{G} & \mathrm{R} \\
 
\hline
 
\end{array}
 
\end{array}</cmath>
 
So, Case (1) has <math>3\cdot6=18</math> ways.
 
  
<u><b>Case (2): The two balls adjacent to the top-left red ball have the same color.</b></u>
+
<u><b>Case (2): The top-center and center-left chips have the same color.</b></u>
<cmath>\begin{tabular}{|c|c|p{2.5 mm}|}
+
<asy>
\hline
+
/* Made by MRENTHUSIASM */
\textbf{R} & \textbf{B} &  \\
+
unitsize(7mm);
\hline
+
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
\textbf{B} &  &  \\
+
fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue);
\hline 
+
fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue);
  &  &  \\
+
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
\hline 
+
draw((6,1)--(9,1), linewidth(1.5));
\end{tabular}</cmath>
+
draw((6,2)--(9,2), linewidth(1.5));
Each placement has <math>6</math> 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).
+
draw((7,0)--(7,3), linewidth(1.5));
 +
draw((8,0)--(8,3), linewidth(1.5));
 +
</asy>
 
There are three sub-cases for Case (2):
 
There are three sub-cases for Case (2):
<cmath>\begin{tabular}{ccccccc}
+
<asy>
\begin{tabular}{|c|c|c|}
+
/* Made by MRENTHUSIASM */
\hline
+
unitsize(7mm);
R & B & R \\ 
+
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
\hline
+
fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue);
B & G & B \\   
+
fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue);
\hline
+
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, green);
G & R & G \\
+
fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red);
\hline
+
fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue);
\end{tabular}
+
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green);
& & &
+
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green);
\begin{tabular}{|c|c|c|}
+
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red);
\hline
+
 
R & B & G \\ 
+
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
\hline
+
fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue);
B & G & R \\
+
fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue);
\hline
+
fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, green);
R & B & G \\
+
fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green);
\hline
+
fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, red);
\end{tabular}
+
fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green);
& & &
+
fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red);
\begin{tabular}{|c|c|c|}
+
fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue);
\hline
+
 
R & B & G \\ 
+
fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red);
\hline
+
fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue);
B & G & R \\   
+
fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, blue);
\hline
+
fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, green);
G & R & B \\
+
fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green);
\hline
+
fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, red);
\end{tabular}
+
fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, blue);
\end{tabular}</cmath>
+
fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, green);
So, Case (2) has <math>3\cdot6=18</math> ways.
+
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 <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.
  
 
<u><b>Answer</b></u>
 
<u><b>Answer</b></u>
Line 157: Line 190:
  
  
<b>Case (2)</b>: All of the rows have two balls 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>.
+
<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>
 
Therefore, our answer is <math>24+12=\boxed{\textbf{(E)} ~36}.</math>
  
 
~michaelchang1
 
~michaelchang1
 +
 +
== Solution 4 (Casework on the Red Chips' Configurations) ==
 +
We will consider all possible configurations of the red chips in 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 <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)
 +
 
== Video Solution (Easiest) ==
 
== Video Solution (Easiest) ==
 
https://www.youtube.com/watch?v=UPUrYN1YuVA ~ MathEx
 
https://www.youtube.com/watch?v=UPUrYN1YuVA ~ MathEx

Latest revision as of 11:18, 25 April 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

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 $\boxed{\text{E}}$.

-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 sub-cases 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 sub-case 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 sub-cases 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 sub-case 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 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. $3\cdot2\cdot2=12$.

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

~michaelchang1

Solution 4 (Casework on the Red Chips' Configurations)

We will consider all possible configurations of the red chips in 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 Flame Kho)

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

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AMC 10 Problems and Solutions

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