To determine the number of ways to arrange $n$ rooks on an $n \times n$ chessboard such that no two rooks attack each other and the placement is symmetric about the main diagonal, let's analyze the problem systematically.

Step 1: Understanding the Constraint

• A valid arrangement means that no two rooks share the same row or column.
• The placement must be symmetric about the diagonal, meaning if there is a rook at $(i, j)$, then there must also be a rook at $(j, i)$.
• This means each valid arrangement corresponds to a permutation matrix that is self-inverse, meaning the permutation must be its own inverse.

Step 2: Counting Such Permutations

• The problem reduces to counting involution permutations (permutations that are their own inverse) in the symmetric group $S_n$.
• Such permutations consist of fixed points (elements mapping to themselves) and 2-cycles (pairs swapping with each other).

The number of such involution permutations in $S_n$ is given by the nth telephone number, denoted by $T_n$, which follows the recurrence relation: $T_n = T_{n-1} + (n-1) T_{n-2}$ with base cases: $T_0 = 1, \quad T_1 = 1$ where:

• The first term accounts for fixing one element and counting the remaining permutations.
• The second term accounts for forming a 2-cycle and counting the remaining elements.

Conclusion:

Thus, the number of ways to place $n$ non-attacking rooks symmetrically about the diagonal is given by:

$R_n = T_n$

Would you like a worked example for a specific $n$? Let me know!

Related Questions:

1. What are the first few values of $R_n$?
2. How do you derive the recurrence relation for involution permutations?
3. Can you express $R_n$ in terms of a summation formula?
4. What is the asymptotic growth of $R_n$ as $n \to \infty$?
5. How does this problem relate to standard rook placements (without symmetry constraints)?

Tip:

A good way to understand symmetric rook placements is to try small values of $n$ manually and observe patterns!