The question is, for a group $G$, and $|G|=24$. We already know it has exactly $8$ elements whose order is $3$. Then how many subgroups does $G$ have.
Here is what I do:
Since it has $8$ elements whose order is $3$, it has $4$ subgroups whose order is $3$, thus $n_3=4$. It also has $2$ trivial subgroups. And by Sylow's third theorem, it has only one Sylow-subgroup of order $8$. But how can I know the structure of this Sylow-subgroup? And I'm not so sure how to deal with other subgroups. Maybe by director product?