Let $U = \{1, 2, \dots, n\}$, where $n$ is a large positive integer greater than 1000. Let $k$ be a positive integer less than $n$. Let $A, B$ be subsets of $U$ with $|A| = |B| = k$ and $A \cap B = \emptyset$. We say that a permutation of $U$ separates $A$ from $B$ if one of the following is true. - All members of $A$ appear in the permutation before any of the members of $B$. - All members of $B$ appear in the permutation before any of the members of $A$. How many permutations of $U$ separate $A$ from $B$?

To construct a valid permutation of $U$ that separates $A$ from $B$:

1. Choose the positions for $A \cup B$: The total number of elements in $A \cup B$ is $|A| + |B| = 2k$. Out of the $n$ available positions in the permutation, we choose $2k$ positions in $\binom{n}{2k}$ ways.

2. Place elements of $A$ and $B$:

   • Case 1 ($A$ before $B$): All $k$ elements of $A$ must occupy the first $k$ chosen positions (which can be arranged in $k!$ ways), and all $k$ elements of $B$ must occupy the remaining $k$ chosen positions (which can be arranged in $k!$ ways). This gives $(k!)^2$ ways.

   • Case 2 ($B$ before $A$): Similarly, all $k$ elements of $B$ occupy the first $k$ chosen positions, and all $k$ elements of $A$ occupy the last $k$ chosen positions. This gives $(k!)^2$ ways.

   Since $A$ and $B$ are disjoint and non-empty, these two cases are mutually exclusive, yielding a total of $2(k!)^2$ ways to place the elements of $A$ and $B$ in the $2k$ chosen positions.

3. Place the remaining elements: The remaining $n - 2k$ elements of $U \setminus (A \cup B)$ can be arranged in the remaining $n - 2k$ positions in $(n - 2k)!$ ways.

Multiplying these possibilities gives:

Thus, option (D) is the correct answer.