How many subgroups of order 17 does $S_{17}$ have? How many subgroups of order 17 does $S_{17}$ have ?
My attempt :
An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle. 
Number of elements of order 17 in $S_{17}$ is $\frac{17!}{17} = 16!$.
Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.
Hence the number of sylow 17 subgroups would be $\frac{16!}{16} = 15!$
 A: Your answer looks correct.
It is true that there are $16!$ elements of order $17$,  but if this is a homework question a marker might want you to elaborate on why that is.
An alternative (but not necessarily better) proof is as follows:
Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.
For each subgroup $G$ fix $g\in G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.
Write $g=(1,2,x_3,\ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups. 
A: Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is
$(p − 2)!$.
Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators.
Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements
of order p are partitioned according to which p-Sylow subgroup they belong to. We need to
count the number of elements of order exactly p. This is precisely the number of distinct
$p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$
Now  take take $p= 17$ ,then 
Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is
$(17 − 2)!=15!$.
