Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to A. n!/2 B. (n-1)! C. (n-1)!/2 D. n!/(2n)

GATE 2019 · Discrete Mathematics · Graph Connectivity · medium

Answer: C, D -- both (n-1)!/2 and n!/(2n) equal the number of distinct Hamiltonian cycles in K_n.

  1. Fix one vertex to eliminate rotation duplicates: Fix vertex 1. The remaining n-1 vertices can be arranged in (n-1)! ways, each forming a unique sequence 1-v_2-v_3-...-v_n-1.
  2. Divide by 2 for direction symmetry: Each cycle 1-v_2-...-v_n-1 is the same as its reverse 1-v_n-...-v_2-1. Divide by 2: answer = (n-1)!/2.
  3. Verify option D equals option C: n!/(2n) = n*(n-1)!/(2n) = (n-1)!/2. Option D equals option C.