Let ๐บ be an undirected complete graph on ๐‘› vertices, where ๐‘› > 2. Then, theโ€ฆ

2019

Let ๐บ be an undirected complete graph on ๐‘› vertices, where ๐‘› > 2. Then, the number of different Hamiltonian cycles in ๐บ is equal to

Answer: D. \(\frac {(๐‘›โˆ’1)!} {2}\) โ€” Answer: (nโˆ’1)!/2 Step 1: Remove cyclic symmetry by fixing one vertex as the start. The remaining (nโˆ’1) vertices can be arranged in (nโˆ’1)! orders, giving allโ€ฆ

  1. A.

    ๐‘›!

  2. B.

    (๐‘› โˆ’ 1)!

  3. C.

    1

  4. D.

    \(\frac {(๐‘›โˆ’1)!} {2}\)

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Correct answer: D

Answer: (nโˆ’1)!/2

  • Step 1: Remove cyclic symmetry by fixing one vertex as the start. The remaining (nโˆ’1) vertices can be arranged in (nโˆ’1)! orders, giving all possible cyclic orderings up to rotation.

  • Step 2: In an undirected graph, each cyclic ordering and its reverse correspond to the same Hamiltonian cycle, so divide by 2 to account for this reflection symmetry.

  • Conclusion: The number of distinct Hamiltonian cycles in the undirected complete graph on n>2 vertices is (nโˆ’1)!/2.

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