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
6 Solution 4
7 Solution 5 (Casework and Symmetry)
8 Solution 6 (Casework and Derangements)
9 Solution 7 (Rush, Use if you have less than 5 minutes)
10 Video Solution (Easiest)
11 Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
12 Video Solution by The Power of Logic
13 Video Solution by MRENTHUSIASM (English & Chinese)
14 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

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 $1-6$ .

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

There are $3! = 6$ ways to assign colors to the cells. The $3$ cells with the same color can either be cells $1,5,3$ or $2,4,6.$ WLOG say cells $1,5,3$ are red. Then cells $2,4$ or $2,6$ can be the $2$ cells sharing a color. WLOG, say cells $2,4$ are blue. Then cell $6$ must be green. So we get a total of $3 \cdot 2 \cdot 1 \cdot 2 \cdot 2 = 24$ colorings.

$[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); // Cell 1 fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); // Cell 2 fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); // Cell 3 fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); // Cell 4 fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); // Cell 5 fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); // Cell 6 // Draw grid draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,0)--(3,3)); // Label cells (A1 at bottom-left) label("7",(0.5,0.5)); label("8", (1.5,0.5)); label("9",(2.5,0.5)); label("4", (0.5,1.5)); label("5", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); [/asy]$

Now, what about the colorings where the grid is not uniquely determined, even if the top two rows are filled out? In other words, what about the colorings where even if we fill out 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 would have to use two tokens of each color, and the second row of the grid would be some permutation of the colors blue, red, green. There are 6 ways to permute the second row. For every permutation, the $3$ x $2$ grid's coloring is uniquely determined. For example, 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 cell $2$ must be red, cell $3$ must be blue, and cell $1$ must be green (this means the $3$ x $2$ grid's coloring was uniquely determined from the second row's colorings). Finally, there are two options for the last row of the 3x3 grid (cells $7,8,9$ can either be green, red, blue or blue green red, respectively), giving us $6 \cdot 2 = 12$ colorings here.

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

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

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

~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

~MRENTHUSIASM

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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2021 AMC 10A Problems/Problem 25

Contents

Problem

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

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

Solution 3 (Casework on the Red Chips' Configurations)

Solution 4

Solution 4

Solution 5 (Casework and Symmetry)

Solution 6 (Casework and Derangements)

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

Video Solution (Easiest)

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

Video Solution by The Power of Logic

Video Solution by MRENTHUSIASM (English & Chinese)

See Also