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

Revision as of 23:10, 10 August 2025

1 Problem
2 Solution 1 (Casework on the Center's Color Chip's Configurations)
3 Solution 2 (Casework on the Top-Center and Center-Left Chips)
4 Solution 3 (Casework on the Red Chips' Configurations)
5 Solution 4 (Focusing On Top 2 Rows)
6 Solution 5 (Casework and Symmetry)
7 Solution 6 (Casework and Derangements)
8 Solution 7 (Rush, Use if you have less than 5 minutes)
9 Video Solution (Easiest)
10 Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
11 Video Solution by The Power of Logic
12 Video Solution by MRENTHUSIASM (English & Chinese)
13 See Also

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 (Focusing On Top 2 Rows)

$[asy] size(100); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); // Label cells (A1 at bottom-left) label("7",(0.5,0.5)); label("8", (1.5,0.5)); label("9",(2.5,0.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); draw((0,1)--(3,1)--(3,3)--(0,3)--cycle, linewidth(1.5)); [/asy]$

We will first count the colorings where if the top two rows are filled out, the grid is uniquely determined (if we fill in two rows, there is only one possibility for the third row to complete the coloring). This will only happen if there are $3$ cells of one color, $2$ cells of another color, and $1$ cell of the remaining color in the top $3$ x $2$ grid made up of cells $1-6$ . With this, we are left with two tokens of one color and one token of another color for the bottom row, so there must only be $1$ way to fill it up since the two tokens of the same color can't be next to each other.

The $3$ cells with the same color can either be cells $1,5,3$ or $2,4,6.$ The two cells with the same color must be cells $4$ and $6$ , and cell $2$ must be the cell of the other color. So there are $2$ ways to choose which cells share $3$ colors and $1$ way to choose which cells share $2$ and $1$ color. There are $3! = 6$ ways to assign colors to the cells. WLOG say cells $1,5,3$ are red and cells $4,6$ are blue. Then cell $2$ must be green. For the last row of the grid, we are left with $2$ green tokens and a blue token. There is only one way to place them (cells $7$ and $9$ must be green and cell $8$ must be blue), so the grid is uniquely determined by the $3$ x $2$ grid's coloring. We have a total of $2 \cdot 1 \cdot 3! = 12$ colorings for this case.

$[asy] size(100); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); // Cell 1 fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); // Cell 2 fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); // Cell 3 fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); // Cell 4 fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); // Cell 5 fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); // Cell 6 fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); // Cell 7 fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, blue); // Cell 8 fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); // Cell 9 draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("7",(0.5,0.5)); label("8", (1.5,0.5)); label("9",(2.5,0.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); [/asy]$

Now, what about the colorings where even if the top two rows are filled out, the grid is still not uniquely determined? In other words, what about the colorings where even if we fill in the top two rows, there is still more than one possible color combination for the last row?

In this case, the $3$ x $2$ grid made up of the top two rows would have to use two tokens of each color, and the second row would be some permutation of the colors blue, red, green. Then for the last row, we would be left with one token of each color.

There are $3! = 6$ ways to permute the second row. WLOG, say we fill the second row of the grid so that cell $4$ is red, cell $5$ is blue, and cell $6$ is green. Then cells $1,2,3$ can be green, red, blue or blue, green, red, respectively. Finally, cells $7,8,9$ can either be green, red, blue or blue, green, red, respectively. This gives us $6 \cdot 4 = 24$ colorings.

$[asy] size(50); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, green); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, red); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, blue); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("7", (0.5,0.5)); label("8", (1.5,0.5)); label("9", (2.5,0.5)); [/asy]$

$[asy] size(50); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, blue); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, blue); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("7", (0.5,0.5)); label("8", (1.5,0.5)); label("9", (2.5,0.5)); [/asy]$

$[asy] size(50); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, green); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, red); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, blue); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("7", (0.5,0.5)); label("8", (1.5,0.5)); label("9", (2.5,0.5)); [/asy]$

$[asy] size(50); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, blue); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, blue); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("7", (0.5,0.5)); label("8", (1.5,0.5)); label("9", (2.5,0.5)); [/asy]$

So our final answer is $12 + 24 = \boxed{36}.$

~grogg007

Solution 5 (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 6 (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

Solution 7 (Rush, Use if you have less than 5 minutes)

Ignore the center piece. Notice that when you place 3 of the chips, there are $6$ ways, making it inevitable for $2$ ways left for the other 2, so $6$ multiply $2$ is $12$ . Now, notice there are $3$ ways to place a center piece, so our final statement is $12\cdot3=\boxed{\textbf{(E)} ~36}.$

~hashbrown2009

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