34. Let a large cube of $4 \times 4 \times 4$ be composed of 64 unit cubes, and select 16 of these unit cubes to be painted red, such that in each $1 \times 1 \times 4$ small rectangular prism composed of 4 unit cubes, there is exactly 1 red cube. How many different ways are there to select the 16 red cubes? Explain your reasoning.
34. Taking the base of a $4 \times 4 \times 4$ cube as the reference plane, divide the large cube into four layers parallel to the reference plane (each layer being a $4 \times 4 \times 1$ rectangular prism), and number the layers sequentially as $1,2,3,4$. Project each red cube onto the reference plane and fill in the layer number of the red cube in the corresponding grid on the reference plane. This results in a $4 \times 4$ grid, where each cell contains a number from $\{1,2,3,4\}$. 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 requirements of the problem. Therefore, the problem is reduced to determining: How many different 4th-order Latin squares are there? Let $a_{ij} (i, j=1,2,3,4)$ represent the number filled in the cell at the $i$-th row and $j$-th column of the $4 \times 4$ grid. For any permutation $\{a, b, c, d\}$ of $\{1,2,3,4\}$, first consider the Latin squares when $a_{11}=a, a_{12}=a_{21}=b, a_{13}=a_{31}=c, a_{14}=a_{41}=d$ (as shown in the figure). Since $a_{22}$ can be any number from $\{a, c, d\}$, we divide it into 3 cases: (1) When $a_{22}=a$, then $a_{23}$ and $a_{32}$ can only be $d$, and $a_{24}$ and $a_{42}$ can only be $c$. Then $a_{33}$ can be $a$ or $b$. Once $a_{33}$ is chosen, the values of $a_{34}, a_{43}, a_{44}$ are uniquely determined. Therefore, there are exactly 2 Latin squares corresponding to this case. (2) When $a_{22}=c$, $a_{24}$ and $a_{42}$ can only be $a$, thus $a_{23}$ and $a_{32}$ can only be $d$, and then $a_{34}$ and $a_{43}$ can only be $b$, and $a_{44}$ can only be $c$, and $a_{33}$ can only be $a$. Therefore, there is exactly 1 Latin square corresponding to this case. (3) When $a_{22}=d$, $a_{24}$ and $a_{42}$ can only be $c$, thus $a_{23}$ and $a_{32}$ can only be $a$, and then $a_{34}$ and $a_{43}$ can only be $b$, and finally, $a_{33}$ can only be $d$, and $a_{44}$ can only be $a$. Therefore, there is exactly 1 Latin square corresponding to this case. In summary, there are exactly 4 Latin squares as shown in the figure. Since there are $4!$ permutations of $\{a, b, c, d\}$, and for a fixed $\{a, b, c, d\}$, the Latin squares shown in the figure can form $3!$ different Latin squares by permuting the 2nd, 3rd, and 4th rows. Therefore, the total number of 4th-order Latin squares is $4! \cdot 3! \cdot 4 = (4!)^2 = 576$. Thus, the total number of ways to select 16 red unit 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 previous solution, the problem can be reduced to finding the number of different $n$-order Latin squares $L_n$. Typically, 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 $l_n$, so we have $L_n = n!((n-1)!) l_n$. As $n$ increases, $l_n$ and $L_n$ grow rapidly. The known results are $l_1=1, l_2=1, l_3=1, l_4=4, l_5=56, l_6=9408, l_7=16942080, \cdots$. For $n \geq 5$, many cases need to be considered to count $l_n$ without repetition or omission, which is impossible to do in a short time. Therefore, choosing $n=4$ as a test question is appropriate.