【Question 12】Cut a $4 \times 4 \times 4$ cube into 64 $1 \times 1 \times 1$ small cubes, then dye 16 of the $1 \times 1 \times 1$ small cubes red, requiring that among any 4 small cubes parallel to any edge, exactly 1 small cube is dyed red. The number of different coloring methods is $\qquad$ (coloring methods that are the same after rotation are also considered different coloring methods).
【Analysis and Solution】
A $4 \times 4 \times 4$ cube, from top to bottom, has 4 layers, each layer being a $4 \times 4 \times 1$ rectangular prism; according to the problem statement "in any 4 small cubes parallel to an edge, exactly 1 small cube is painted red", in each layer of $4 \times 4 \times 1$, there are exactly 4 small cubes painted red;
and when viewed from above, the top view of the small cubes painted red in each layer is a $4 \times 4$ square.
For this $4 \times 4$ square, we use “1”, “2”, “3”, “4” each 4 times,
to represent the small squares in the 1st, 2nd, 3rd, and 4th layers that are painted red;
and according to the problem, in this $4 \times 4$ square, each row and each column contains “1”, “2”, “3”, “4” exactly once.
Thus, the problem is transformed into “how many different ways are there to fill a $4 \times 4$ square with “1”, “2”, “3”, “4” each 4 times, such that each row and each column contains “1”, “2”, “3”, “4” exactly once?”
We first consider “1”, with one in each row and in different columns, there are $4 \times 3 \times 2 \times 1 = 24$ ways to fill; let's assume it is in the configuration of Figure 1.
Next, we consider “2”, the “2” in the first row has 3 ways to fill; let's assume it is in the configuration of Figure 2;
if the “2” in the second row is in the first column, then the “2” in the third row is in the fourth column, and the “2” in the fourth row is in the third column; (as in Figure 3)
if the “2” in the second row is in the third column, then the “2” in the fourth row is in the first column, and the “2” in the third row is in the fourth column; (as in Figure 4)
if the “2” in the second row is in the fourth column, then the “2” in the third row is in the first column, and the “2” in the fourth row is in the third column; (as in Figure 5)
Next, we consider “3”, the “3” in the first row has 2 ways to fill; let's assume the “3” in the first row is in the third column (as in Figures 6 to 8)

Figure 6 has 2 remaining ways to fill; (as in Figures 9 and 10)
Figure 7 has 1 remaining way to fill; (as in Figure 11)
Figure 8 has 1 remaining way to fill; (as in Figure 12)
Thus, if the “2” in the first row is in the second column and the “3” in the first row is in the third column, there are 4 ways;
the total number of different ways to fill is $24 \times 3 \times 2 \times 4 = 576$.
Therefore, the number of different coloring methods for the original problem is 576.