Six, let a large cube of $4 \times 4 \times 4$ be composed of 64 unit cubes. Select 16 of these unit cubes to be painted red, such that in the large cube, each $1 \times 1 \times 4$ small rectangular prism composed of 4 unit cubes contains exactly 1 red cube. How many different ways are there to select the 16 red cubes? Explain your reasoning.
Solution: Choose a $4 \times 4$ side of the $4 \times 4 \times 4$ large cube as the base plane. Divide the large cube into four layers parallel to the base plane (each layer being $4 \times 4 \times 1$ in size), and number these layers as $1, 2, 3, 4$. Project each red unit cube onto the base plane and write the layer number of the red unit cube in the corresponding grid. This way, we obtain a $4 \times 4$ grid, where each cell contains a number from $\{1, 2, 3, 4\}$, and cells with the same number are neither in the same row nor in the same column. Such a $4 \times 4$ grid is called a 4th-order Latin square. Conversely, each 4th-order Latin square uniquely determines a coloring method that meets the problem's requirements, and different Latin squares determine different coloring methods. Therefore, the problem is reduced to determining: how many different 4th-order Latin squares are there.

We agree to label the positions of the individual cells in the $4 \times 4$ grid as shown in Figure 1. For any permutation $(a, b, c, d)$ of $(1, 2, 3, 4)$, we will examine the number of Latin squares as shown in Figure 2. For this purpose, we first examine the number of Latin squares as shown in Figure 3.

The $(2,2)$ cell in Figure 3 can be filled with any number from $\{a, c, d\}$. We will examine these three cases separately.
(i) Suppose the $(2,2)$ cell in Figure 3 is filled with $a$.
In this case, the $(2,3)$ and $(3,2)$ cells can only be filled with $d$; then the $(2,4)$ and $(4,2)$ cells can only be filled with $c$; then the $(3,3)$ cell can be filled with $a$ or $b$. Once the $(3,3)$ cell is filled, the remaining three cells are uniquely determined. Therefore, there are exactly 2 Latin squares corresponding to this case.
(ii) Suppose the $(2,2)$ cell in Figure 3 is filled with $c$.
In this case, the $(2,4)$ and $(4,2)$ cells can only be filled with $a$; then the $(2,3)$ and $(3,2)$ cells can only be filled with $d$; then the $(3,4)$ and $(4,3)$ cells can be filled with $b$. Once these cells are filled, the remaining cells are uniquely determined. Therefore, there is exactly 1 Latin square corresponding to this case.
(iii) Suppose the $(2,2)$ cell in Figure 3 is filled with $d$.
Similar to case (ii), there is only one unique filling method for this case, i.e., there is exactly 1 Latin square.

Based on the above discussion, there are exactly 4 Latin squares as shown in Figure 3. Each such Latin square can generate $3!$ different Latin squares as shown in Figure 2 by permuting the 2nd, 3rd, and 4th rows. Therefore, there are exactly $4 \times 3!$ Latin squares as shown in Figure 2. Finally, since there are $4!$ permutations of $(a, b, c, d)$, the total number of 4th-order Latin squares is $(4!)^2$.

Thus, the total number of ways to select 16 red cubes is $(4!)^2 = 576$.

Note: Consider a more general problem, in an $n \times n \times n$ large cube, select $n^2$ unit cubes to be painted red, such that any $1 \times 1 \times n$ rectangular prism composed of $n$ unit cubes contains exactly one red unit cube. How many different ways are there to select $n^2$ red unit cubes that meet the above requirements?

Following the solution above, the problem can be reduced to finding the number of different $n$-order Latin squares $L_n$. The number of $n$-order Latin squares with the first row and first column being the standard permutation $(1, 2, \cdots, n)$ is denoted as $l_n$. Thus,
$$
L_n = (n!)((n-1)!) l_n.
$$

As $n$ increases, $l_n$ and $L_n$ grow rapidly. People have only calculated $l_n$ and $L_n$ for smaller $n$.
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|}
\hline $n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline $l_n$ & 1 & 1 & 1 & 4 & 56 & 9408 & 16942080 & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}

For $n \geq 5$, a large number of cases must be analyzed to accurately calculate the large number $l_n$ without repetition or omission, which is difficult to achieve in a short time. Therefore, choosing the case $n=4$ can be considered appropriate.
(Provided by the Examination Committee, written by Zhang Zhusheng)