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!
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.
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 !