Sylow subgroups of A6 ($n_5=36,n_3=10$)

In the alternating group $A_6$, of order $360$, a Sylow subgroup of order five is generated by a five-cycle. There are $6\cdot 4!=144$ five-cycles, four in each such subgroup, giving $36$ subgroups. A Sylow subgroup of order nine is generated by disjoint three-cycles, hence is $C_3\times C_3$. The unordered partitions of six letters into two triples give all such subgroups by Sylow conjugacy, and their single three-cycles recover the partition uniquely. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions.