AMC10 2023 A

Let a "tile" denote a 1×1 square, and a "square" refer to a 2×2 square. We have 4!=24 possible ways to fill out the top-left square. Next, we fill out the bottom-right corner tile. In the bottom-right square, one corner is already filled (the central tile) from our initial coloring), so we have 3 color options remaining for this. Now considering the remaining tiles, all of these only have one way to be filled (Try it yourself if you don’t believe). For example, the right tile of the middle row is part of two squares: the top-right and the bottom-right. Among these squares, 3 colors have already been used, leaving us with only 1 remaining option. Similarly, every other remaining square has only one available option for coloring. Thus, the total number of ways is 3×4!=(D) 72. A quick version of this method is used when you use for example B as the top left corner block. You can see that $B$ in the top left corner has $6$ possible ways. Now, see that there are $3$ possible corners for a center cell and there are $4$ possible center cells. We get $6$ times $4$ times $3 = D$, the answer is (D) 72.