Can someone explain to me why the symmetric group EDIT: $S_6$ has 10 Sylow 3-subgroups? I gather the Sylow 3-subgroups of $S_6$ are isomorphic to $\mathbb Z_3 \times\mathbb Z_3$ and that it must be of order 9, by prime decomposition. I thought I could use the binomial formula $ {6 \choose 3}$ to work out the number of sylow 3-subgroups, but of course this gives me 20.
I realise this is probably a simple question but any help would be appreciated.
The $3$-Sylow subgroups are generated by pairs of disjoint $3$-cycles (this is where the $\mathbb{Z}_3 \times \mathbb{Z}_3$ comes from). How many ways can you choose a pair of disjoint $3$-cycles? You choose a $3$-cycle, and then its partner is already determined. So far there are $\binom{6}{3}$ ways to do this.
However, we have double counted. If instead, we had chosen our cycle's partner, then we would get the same two generators. This is why you divide by $2$.