2023 AMC 10A Problem 20

Label the cells row by row as $a,$ $b,$ $c;$ $d,$ $e,$ $f;$ $g,$ $h,$ $i;$ the top-left four are a permutation of all four colors

Then $\{c, f\}$ and $\{g, h\}$ are each the two remaining colors in some order, and $i$ is forced but works only when $f \ne h$

Label the cells row by row as $a,$ $b,$ $c;$ $d,$ $e,$ $f;$ $g,$ $h,$ $i.$ 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.