How many Sylow p-subgroups are in this group 
By direct computation, find the number of Sylow $3$-subgroups and the number of Sylow $5$-subgroups of the symmetric group $S_5$.
Solution: In $S_5$ there are $20$ three cycles which split up into $10$ subgroups of order $3$. This number is congruent to $1 \mod 3$ and is a divisor of ($5 \cdot 4 \cdot 2$).
There are $24$ five cycles which split up into $6$ subgroups of order $5$. This number is congruent to $1 \mod 5$ and is a divisor of ($5 \cdot 4 \cdot 2$).

My question is where did the numbers $10$ subgroups and $6$ subgroups come from?

A (cyclic) group of order 3 is always of the form 
$$\{e,a,a^{-1}\} $$
where $e$ is the identity, and $a$ and $a^{-1}$ have order 3. Thus the number $10$ is the result of the calculation $$\frac{20}{3-1}=\frac{20}{2}$$
(the identity element is not of order 3). Similarly
$$6=\frac{24}{5-1}. $$
