2021 AMC 10A Problems/Problem 25

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 Solution 8 (Casework Using the Center and Surrounding)
10 Simple Video Solution by grogg007 (2 cases)
11 Video Solution (Easiest)
12 Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
13 Video Solution by The Power of Logic
14 Video Solution by MRENTHUSIASM (English & Chinese)
15 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)

$[asy] size(50); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); // Label cells (A1 at bottom-left) label("7",(0.5,0.5)); label("8", (1.5,0.5)); label("9",(2.5,0.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); draw((0,1)--(3,1)--(3,3)--(0,3)--cycle, linewidth(1.5)); [/asy]$

We will first count the colorings where if the top two rows are filled out, the grid is uniquely determined. 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. This makes the grid uniquely determined.

The $3$ cells with the same color can either be cells $1,5,3$ or $2,4,6.$ If we choose $1,3,5,$ the two cells with the same color must be cells $4$ and $6$ and cell $2$ must be the cell of the other color. If we choose $2,4,6,$ the two cells with the same color must be cells $1$ and $3,$ and cell $5$ 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. We have a total of $2 \cdot 1 \cdot 3! = 12$ colorings for this case. An example coloring is shown below with cells $1,3,5$ being chosen for the $3$ cells of the same color.

$[asy] size(70); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); // Cell 1 fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); // Cell 2 fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); // Cell 3 fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); // Cell 4 fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); // Cell 5 fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); // Cell 6 fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); // Cell 7 fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, blue); // Cell 8 fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); // Cell 9 draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); label("7",(0.5,0.5)); label("8", (1.5,0.5)); label("9",(2.5,0.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("Only 1 configuration for last row, grid is uniquely determined from top 2 rows", (1.5,-1.2)); [/asy]$

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 also either be green, red, blue or blue, green, red, respectively. This gives us $6 \cdot 4 = 24$ colorings.

$[asy] size(200); void drawGrid(pair offset, pen[] colors) { fill(shift(offset)*((0,2)--(1,2)--(1,3)--(0,3)--cycle), colors[0]); fill(shift(offset)*((1,2)--(2,2)--(2,3)--(1,3)--cycle), colors[1]); fill(shift(offset)*((2,2)--(3,2)--(3,3)--(2,3)--cycle), colors[2]); fill(shift(offset)*((0,1)--(1,1)--(1,2)--(0,2)--cycle), colors[3]); fill(shift(offset)*((1,1)--(2,1)--(2,2)--(1,2)--cycle), colors[4]); fill(shift(offset)*((2,1)--(3,1)--(3,2)--(2,2)--cycle), colors[5]); fill(shift(offset)*((0,0)--(1,0)--(1,1)--(0,1)--cycle), colors[6]); fill(shift(offset)*((1,0)--(2,0)--(2,1)--(1,1)--cycle), colors[7]); fill(shift(offset)*((2,0)--(3,0)--(3,1)--(2,1)--cycle), colors[8]); for (int i = 0; i <= 3; ++i) { draw(shift(offset)*((i,0)--(i,3))); draw(shift(offset)*((0,i)--(3,i))); } label("1", offset + (0.5,2.5)); label("2", offset + (1.5,2.5)); label("3", offset + (2.5,2.5)); label("4", offset + (0.5,1.5)); label("5", offset + (1.5,1.5)); label("6", offset + (2.5,1.5)); label("7", offset + (0.5,0.5)); label("8", offset + (1.5,0.5)); label("9", offset + (2.5,0.5)); } pen[] colors1 = {green, red, blue, red, blue, green, green, red, blue}; pen[] colors2 = {blue, green, red, red, blue, green, green, red, blue}; pen[] colors3 = {green, red, blue, red, blue, green, blue, green, red}; pen[] colors4 = {blue, green, red, red, blue, green, blue, green, red}; drawGrid((0,0), colors1); drawGrid((4,0), colors2); drawGrid((8,0), colors3); drawGrid((12,0), colors4); [/asy]$

Our final answer is $12 + 24 = \boxed{\textbf{(E)} ~36}$ .

~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

Solution 8 (Casework Using the Center and Surrounding)

There are 3 indistinguishable red chips, 3 indistinguishable blue chips, and 3 indistinguishable green chips. Let's label the red chips R, the blue chips B, and the green chips G.

WLOG, let's assume that the chip in the middle of the array is R. $\begin{tabular}{ c c c } ? & ? & ? \\ ? & R & ? \\ ? & ? & ? \end{tabular}$ For the convenience of our proof, we will replace the question marks as follows. $\begin{tabular}{ c c c } 1 & 2 & 3 \\ 4 & R & 6 \\ 7 & 8 & 9 \end{tabular}$ Doing this helps us easily identify where to place a certain chip. For example, if I want to place a green chip in the top leftmost square, I'd say, "Place a green chip in position 1."

Now that we've assumed the center square is occupied by a red (R) chip, notice that the chips in positions $2, 4, 6, and 8$ can only be blue or green. With the information we now know, we can break the problem into cases based off of the number of a certain color of chip in the positions $2, 4, 6, 8$ .

Case 1 (2 consecutive blue, 2 consecutive green): An example case to show you what this means is shown below. $\begin{tabular}{ c c c } 1 & G & 3 \\ B & R & G \\ 7 & B & 9 \end{tabular}$ There are $4$ ways to figure out where the two blue and two green chips are placed. Notice that after I've placed the two blue, two green, and one red chip, the rest of the chips are in fixed places (try it!). Now we only need to worry about what color the center chip is.

There are 3 ways to choose the color for the center chip, and afterwards, there are only two colors left for the two kinds of chips in positions $2, 4, 6,$ and $8$ , and there are two of each, so there is no need to worry about the color of those chips.

In conclusion, there are $4*3 = 12$ ways to arrange the chips in this case.

Case 2 (3 blue, 1 green OR 3 green, 1 blue): An example case to show you what this means is shown below. $\begin{tabular}{ c c c } 1 & G & 3 \\ B & R & G \\ 7 & G & 9 \end{tabular}$ WLOG, let's first assume that there are $3$ green chips in such a position, and only $1$ blue chip. There are $4$ ways to figure out where they are placed. Notice that after I've placed the three green, one blue, and one red chip, the rest of the chips are yet again in fixed places (try it!). Now we only need to worry about what color the center chip is and which color of chip has $3$ in the positions of $2, 4, 6, 8$ .

There are $3$ ways to choose the color of the center chip. After that is chosen, there are only two colors left for the two kinds of chips needed in the positions $2, 4, 6,$ and $8$ . However, since this time there are a different number of each color of chip, there are now $2$ ways to assign colors to the chips.

In conclusion, there are $4*3*2 = 24$ ways to arrange the chips in this case.

Notice how there should be a Case 3 with cases that look like some variation of this: $\begin{tabular}{ c c c } 1 & G & 3 \\ B & R & B \\ 7 & G & 9 \end{tabular}$ There isn't, because if such an arrangement were to happen, we wouldn't be able to arrange the grid following the problem's conditions. There would either be a blue chip next to a blue chip, and/or a green chip next to a green chip.

$12+24+0 = 36$ , so the answer is $\boxed{\textbf{(E)} ~36}$ .

~ AVRILAVIGNE

Simple Video Solution by grogg007 (2 cases)

https://www.youtube.com/watch?v=L5kI8cTAzyI

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