Counting solutions of $x^{30}=1$ in $S_6$ 
I was asked to count the number of solution of $$x^{30} =1$$ with $x \in S_{6}$     (the group of the permutations over a set with $6$ elements).

I started noticing that the divisors of $30$ are: $1,2,3,5,6,10,15,30$ so I have to count all the permutations of order one divisor of $30$. During the counting I use the fact that every permutations has an unique factorization in disjoint cycles.
So I'm starting counting the number of permutations of order $2$. 
We have $15$ different $ 2 -$cycle, $45$ way to combine $2$ disjoint $ 2-$cycles and $15$ way to combine $3$ disjoint $2-$cycles. We don't have any other elements of,order $2$ in $S_6$.
Order $3$ is similar, we have $40$ different $3-$cycles and $40$ different products of $2$ disjoint $3-$cycles.
Of order $5$ we have only the $5-$cycles ( $144$) and nothing else because we are in $S_6$.
Of order $6$ we have $120$ different $6-$cycles and $120$ way to compose a $3-$cycle with a disjoint $2-$cycle. 
I didn't found elements of order $10,15,30$ because we need at least $S_7$ to have an element of order $10$ ( a $2-$cycle composed with a  disjoint $5-$cycle)
So in total I counted $ 539+1=540$ solution of the equation. To count every case I use a combinatorial approach.
Is my reasoning correct? Is there a faster (and more group-related let's say) way to count the solutions in this case?
I thought about the orbit-stabilizer theorem but I had problems calculating the size of the stabilizer of every conjugation class.
Thanks in advance :)
NOTE: I've redone all my calculations and now are correct :) still the best methods are the ones,explained in the answers
 A: [Disclaimer: The following is a complete solution to the question. Continue reading if you're looking for a complete solution.]
We have $x^{30}\ne 1$ iff $x$ contains a cycle whose length does not divide $30$. Possible cycle lengths are $1$, $\ldots$, $6$, among which only $4$ does not divide $30$.
You can choose a four-cycle in ${6\cdot5\cdot4\cdot3\over4}=90$ ways, and the remaining two elements can either stay fixed or be combined to a $2$-cycle. It follows that there are $180$ permutations in $S_6$ with $x^{30}\ne1$; therefore the remaining $540$ permutations $x$ satisfy $x^{30}=1$.
A: It is simplier to count the number of solutions of $x^{30}\ne 1$. Since the order of a permutation equals to lcm of lengths of its cycles, then $x$ must be of the form either $(....)$ or  $(..)(....)$ (then it orders is divided by $4$). Their number is $2\cdot С^4_6\cdot P_3$ (here $C^m_n$ is the  binomial coefficient, $P_n$ is the number of permutations).
A: You can find the number of solutions machinery. GAP helps to find it as follows:
 > S6:=SymmetricGroup(6);;
 > Size(Union(Filtered(S6,x-> Order(x)=1),Filtered(S6,x-> Order(x)=2),Filtered(S6,x-> Order(x)=3),Filtered(S6,x-> Order(x)=6),Filtered(S6,x-> Order(x)=5)));

 > 540


Regarding to @Alexander's neat comment: you can do it also by:
 > S6:=SymmetricGroup(6);;
 > Size(Filtered(S6,x-> Order(x) in [1,2,3,5,6]));
 > 540

or
 > Number(Filtered(S6,x-> Order(x) in [1,2,3,5,6]));      
 > 540

