AMC 10 Step by Step
Log inSign up
Printable

2021 AMC 10A Problem 25

Problem onlyBack to the problem
2021 AMC 10A Problem 25AMC 10 Step by Step
Name:                             Date: September 13, 2026
Paths & Grids · difficulty 4 of 5 · about 9 min · position P21-25

How many ways are there to place 33 indistinguishable red chips, 33 indistinguishable blue chips, and 33 indistinguishable green chips in the squares of a 3×33 \times 3 grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?

  1. A)

    1212

  2. B)

    1818

  3. C)

    2424

  4. D)

    3030

  5. E)

    3636

Answer

E

Key insight

The center touches every edge square, so its color's other two chips sit in corners; the leftover squares form paths that the remaining two colors must color alternately.

Solution

Suppose the center chip is red (multiply by 33 at the end for the center's color). The center is adjacent to all four edge-middle squares, so the other two red chips must occupy two of the four corners; corners are never adjacent to each other, so any pair of corners is allowed.

Case 1: the two red corners share a side of the grid (4 choices, e.g. top-left and top-right). The top-middle square is then surrounded by red on all three sides and is isolated. The other five empty squares, left-middle, bottom-left, bottom-middle, bottom-right, right-middle, form a path of length 55. Coloring a path with blue and green so no neighbors match means alternating, so the path is XYXYXXYXYX with 33 of color XX and 22 of YY; the isolated square takes the third YY. Choosing XX gives 22 colorings. Subtotal 4⋅2=84 \cdot 2 = 8.

Case 2: the two red corners are opposite (2 choices, e.g. top-left and bottom-right). The empty squares split into two paths of length 33: top-middle, top-right, right-middle; and left-middle, bottom-left, bottom-middle. Each path alternates as XYXXYX; to use exactly three blues and three greens, one path is BGBBGB and the other GBGGBG: 22 colorings. Subtotal 2⋅2=42 \cdot 2 = 4.

With a red center there are 8+4=128 + 4 = 12 arrangements, so in total 3⋅12=363 \cdot 12 = 36.

The answer is (E) 36\boxed{\textbf{(E)}\ 36}.

Why this works

Start with the most connected square: the center's color is forced into the corners, and removing those squares breaks the grid into paths, where a two-coloring with no adjacent repeats is rigid (determined by one square). The only bookkeeping left is the count of each color, which distinguishes the two cases. Reducing a grid problem to paths is a recurring trick for adjacency constraints.

The trap

Treating the arrangement as three rows each needing a permutation of the colors (which ignores vertical adjacency), or forgetting to multiply by 3 for the center's color.

Common mistakes

  • Treating the arrangement as three rows each needing a permutation of the colors (which ignores vertical adjacency), or forgetting to multiply by 3 for the center's color.
  • In Case 1, giving the isolated square the color that already appears three times on the path, which leaves color counts of 4 and 2 instead of 3 and 3.

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects