Question

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?

$12$

$18$

$24$

$30$

$36$

Solution:

Let the colors be $A, B,$ and $C.$ Note that we can assign the $3$ colors to them in $3! = 6$ ways, so we have to multiply by $6$ at the end.

Let $A$ be in the center of the grid.

$\begin{array}{ccc} ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{array}$

The other $A$ s have to either be along the diagonal or on the same side.

$\begin{array}{ccc} ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{array}$

$\begin{array}{ccc} A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{array}$

The first scenario can happen in $2$ ways since there are $2$ diagonals. The second has $4$ ways since there are $4$ sides.

Either way, the positions of the $B$ s and $C$ s is fixed.

$\begin{array}{ccc} C & B & A \\ B & A & C \\ A & C & B \end{array}$

$\begin{array}{ccc} A & B & A \\ C & A & C \\ B & C & B \end{array}$

This is a total of $4 + 2 = 6$ ways to arrange the $A, B,$ and $C$ s. This gives us a total of $6 \cdot 6 = 36$ configurations.

Thus, E is the correct answer.

2021 AMC 10A Spring Problem 25

Solution:

Problem 25 in Other Years