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2021 AMC 10A Problems/Problem 25

Revision as of 20:39, 22 February 2021 by MRENTHUSIASM (talk | contribs) (→‎Problem: The problem should end in a question mark.)

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)

Without the loss of generality, we place a red ball in the top-left square. There are two cases:

Case (1): The two balls adjacent to the top-left red ball have different colors. \[\begin{tabular}{|c|c|p{2.5 mm}|} \hline R & B & \\ \hline G & R & \\ \hline & & \\ \hline \end{tabular}\] Each placement has $6$ permutations, as there are $3!=6$ ways to permute RBG.

There are three sub-cases for Case (1): \[\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline G & R & G \\ \hline B & G & B \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline G & R & B \\ \hline B & G & R \\ \hline \end{tabular} \end{tabular}\] So, Case (1) has $3\cdot6=18$ ways.


Case (2): The two balls adjacent to the top-left red ball have the same color. \[\begin{tabular}{|c|c|p{2.5 mm}|} \hline R & B & \\ \hline B & & \\ \hline & & \\ \hline \end{tabular}\] Each placement has $6$ permutations, as there are $\binom32\binom21=6$ ways to choose three balls consisting of exactly two colors (RBB, RGG, BRR, BGG, GRR, GBB). There are three sub-cases for Case (2): \[\begin{tabular}{ccccccc} \begin{tabular}{|c|c|c|} \hline R & B & R \\ \hline B & G & B \\ \hline G & R & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline R & B & G \\ \hline \end{tabular} & & & \begin{tabular}{|c|c|c|} \hline R & B & G \\ \hline B & G & R \\ \hline G & R & B \\ \hline \end{tabular} \end{tabular}\] So, Case (2) has $3\cdot6=18$ ways.


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 balls 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

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

https://youtu.be/wKJ9ppI-8Ew

~ pi_is_3.14

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 10 Problems and Solutions

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