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I understand the answer that we can choose the 3 cycle in nC3 ways, but why is there 2 ways to do this? Hence times 2?

Thanks!

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If i j and k are the chosen elements moved by the cycle, then (ijk) and (ikj) are both 3 cycles on these letters which are different, and there are no others.

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Let's say the $3$ elements in the $3$ cycle are $a,b,c$ then there are $6$ was to permute these but each perm will be equal to $2$ others \begin{eqnarray*} (abc)=(bca)=(cab) \\ (acb)=(cba)=(bac). \end{eqnarray*} $2$ ways !

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