Question

Each square in a $3 \times 3$ grid of squares is colored red, white, blue, or green so that every $2 \times 2$ square contains one square of each color. One such coloring is shown below (letters denote the colors, with the center square white). How many different colorings are possible?

$24$

$48$

$60$

$72$

$96$

Solution:

Label the cells row by row $a, b, c / d, e, f / g, h, i.$ The top-left block $a, b, d, e$ is a permutation of the four colors, so $4! = 24$ ways. The block $\{b, c, e, f\}$ is also all four colors, and $b, e$ are fixed, so $\{c, f\}$ is the remaining two in some order: $2$ ways. Same story for $\{g, h\},$ the two colors apart from $d, e,$ another $2$ ways. That leaves $i,$ forced to whatever color is missing from $\{e, f, h\},$ and that only works when $f \ne h.$ Of the $2 \cdot 2 = 4$ order combinations, exactly one has $f = h,$ so $3$ survive. The total is $24 \cdot 3 = 72.$ Therefore, the answer is D.

2023 AMC 10A Problem 20

Solution:

Problem 20 in Other Years