Question

Let \(U = \{1,2,\ldots,n\}\) with \(n > 1000\). Let \(k < n\) be a positive integer. Let \(A,B \subseteq U\) with \(|A| = |B| = k\) and \(A \cap B = \varnothing\). A permutation of \(U\) separates \(A\) from \(B\) if either all members of \(A\) appear before any member of \(B\), or all members of \(B\) appear before any member of \(A\).

How many permutations of \(U\) separate \(A\) from \(B\)?

• A. $n!$
• B. $\binom{n}{2k}(n-2k)!$
• C. $\binom{n}{2k}(n-2k)!\,(k!)^2$
• D. $2\binom{n}{2k}(n-2k)!\,(k!)^2$

Answer 1

The Correct Option is A

The problem requires us to calculate the number of permutations of the set \( U = \{1,2,\ldots,n\} \) that separate subsets \( A \) and \( B \). Given that \( |A| = |B| = k \) and \( A \cap B = \varnothing \), we need to understand the condition under which the separation occurs. A permutation separates \( A \) from \( B \) if either all members of \( A \) precede all members of \( B \), or vice versa.

To solve this, consider any permutation of the \( n \) elements. Out of all possible permutations, each provides a unique ordering of elements. For a permutation to separate \( A \) from \( B \), the block of elements represented by \( A \) must appear entirely before or after the block represented by \( B \). Since arrangement within \( A \) and arrangement within \( B \) can occur freely within their respective groupings, the total ways to arrange all elements is simply: \[ n! \]

This is because any permutation of \( U \) will naturally separate \( A \) from \( B \) when viewed appropriately (either \( A \) is entirely before \( B \) or vice-versa). Therefore, no additional constraints beyond permuting all elements are required.

Thus, the number of permutations of \( U \) that separate \( A \) from \( B \) is: \(\boxed{n!}\)