# Number of Hamilton Cycles in Complete Graph

## Theorem

For all $n \ge 3$, the number of distinct Hamilton cycles in the complete graph $K_n$ is $\dfrac {\left({n-1}\right)!} 2$.

## Proof

In a complete graph, every vertex is adjacent to every other vertex.

Therefore, if we were to take all the vertices in a complete graph in any order, there will be a path through those vertices in that order.

Joining either end of that path gives us a Hamilton cycle.

From Cardinality of Set of Bijections, there are $n!$ different ways of picking the vertices of $G$ in some order.

Hence there are $n!$ ways of building such a Hamilton cycle.

Not all these are different, though.

On any such cycle, there are:

• $n$ different places you can start;
• $2$ different directions you can travel.

So any one of these $n!$ cycles is in a set of $2n$ cycles which all contain the same set of edges.

So there are $\dfrac {n!} {2n} = \dfrac {\left({n-1}\right)!} 2$ distinct Hamilton cycles.

$\blacksquare$