In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?


Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $n\ge 3$, but it is less than $2n$ for $n\le 2$.

  • $\begingroup$ I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path. $\endgroup$ – Bob Krueger Dec 19 '18 at 8:50
  • $\begingroup$ @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too. $\endgroup$ – Alex Ravsky Dec 21 '18 at 7:02
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    $\begingroup$ Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle. $\endgroup$ – Bob Krueger Dec 21 '18 at 9:05

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