Math Problem Statement

Recall that a rook in chess attacks along rows and columns. Define R_n to be the number of ways of arranging n rooks on an n x n chess board so that no two can attack each other and their placement is symmetric about the diagonal from lower left to upper right. Which is a correct formula for R_n? R_n = n!/2 R_n = n(n-1)/2 R_n = R_{n-1} + (n-1)R_{n-2} R_n = nR_{n-1} + (n-1)*R_{n-2}

Solution

The problem involves arranging nn rooks on an n×nn \times n chessboard such that their placement is symmetric about the main diagonal (the diagonal from the lower left to the upper right) and no two rooks attack each other. Let's analyze the possible formulas:

Key observations:

  1. Symmetry about the diagonal: If a rook is placed at position (i,j)(i, j), another rook must be placed at (j,i)(j, i) to maintain symmetry.
  2. Non-attacking condition: No two rooks can be in the same row or column.

Understanding the structure:

  • The positions of the rooks can be described as permutations of the set {1,2,,n}\{1, 2, \dots, n\} such that each number appears exactly once in the sequence, representing rows and columns.
  • To ensure diagonal symmetry, the placement must follow the constraint that (i,j)(i, j) is paired with (j,i)(j, i). This means a rook on (i,j)(i, j) forces a placement at (j,i)(j, i).

Recursive analysis:

To compute RnR_n, consider:

  1. Base cases:

    • R1=1R_1 = 1: One rook fits on a 1×11 \times 1 board.
    • R2=1R_2 = 1: There is only one way to place two rooks symmetrically on a 2×22 \times 2 board.
  2. Recursion relation:

    • When placing a rook at position (i,i)(i, i), there are nn choices for ii.
    • The remaining positions reduce the problem to placing n1n-1 rooks on an (n1)×(n1)(n-1) \times (n-1) board.
    • Additionally, off-diagonal placements may pair with (i,j)(i, j) and (j,i)(j, i), introducing terms involving Rn2R_{n-2}.

The correct recursive relation is: Rn=nRn1+(n1)Rn2.R_n = n \cdot R_{n-1} + (n-1) \cdot R_{n-2}.

This corresponds to the fourth formula: Rn=nRn1+(n1)Rn2.R_n = n \cdot R_{n-1} + (n-1) \cdot R_{n-2}.


Final Answer:

The correct formula is: Rn=nRn1+(n1)Rn2.R_n = n \cdot R_{n-1} + (n-1) \cdot R_{n-2}.

Would you like a step-by-step derivation or examples to illustrate this?

Related Questions:

  1. What are the explicit values of R3R_3 and R4R_4 based on the recurrence relation?
  2. How does the diagonal symmetry condition restrict the arrangement of rooks?
  3. Can RnR_n be expressed in terms of factorials or combinatorial coefficients?
  4. What is the growth rate of RnR_n as nn \to \infty?
  5. How does the problem change if we consider symmetry about other diagonals?

Tip:

To solve problems with recursive formulas, always analyze base cases and carefully account for all contributing terms in the recurrence relation.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Symmetry in Chessboard Problems
Recursion

Formulas

R_n = n * R_{n-1} + (n-1) * R_{n-2}

Theorems

Permutation of Rows and Columns
Diagonal Symmetry Constraints

Suitable Grade Level

Grades 10-12 and above (Advanced Mathematics)