Formula for the number of elements in $S_{10}$ of order $10$. I'm trying to find a formula (mostly in terms of factorials) for the number of elements in $S_{10}$ of order $10$.
We first we count how many elements of $S_{10}$ have the same cycle structure.
So,since $lcm(a,b)=10⟺a=10,b=0$ or $a=5,b=2
$, there are only one $10−cycle$ or ,one $5−cycle$, two $2−cycle$, one $1−cycle$, and therefore the number of elements in $S_{10}$ with order $10$ is equal to $\frac{10!}{5∗2∗4}+\frac{10!}{10}=9!+ \frac{9!}{4}$.
Am I right?
 A: Step $1$: any permutation can be written (using the cycle notation) as a product of disjoint cycles (in particular, they commute).
Step $2$: the order of a product of disjoint cycles is the lowest common multiple of the lengths of the cycles. This is fairly easy to prove.
Step $3$: Since $10=2 \cdot 5$, an element in $S_{10}$ has order $10$ if and only if it is composed of a single cycle of length $10$, or of cycles of lengths $5$ and $2$ (minimum one for each). So you get the following types: $$(..........)\quad,\quad (..)(.....)\quad,\quad (..)(..)(.....)$$
Step $4$: Using combinatorics, compute the possibilities for each and add them up.
For a $10$-cycle there are $9!$ possibilities, as by definition $1$ is always the first digit and then you can freely permute the other $9$.
For a $(2,5)$-cycle there are ${10 \choose 3} \cdot {7 \choose 2}\cdot 4!$, as first you choose the three elements that are fixed, then among the remaining seven you choose the ones in the two-cycle, and finally you count how many five-cycles you can form using the five elements that remain.
For a $(2,2,5)$-cycle there are ${10 \choose 1} \cdot {9 \choose 2} \cdot {7 \choose 2} \cdot 4! \cdot \frac{1}{2}$. The logic is the same as above, but in the end you have to divide by two to account for double counting of the two cycles (with the process above you're counting $(12)(34)$ and $(34)(12)$ as separate elements)
A: For explanation see the answers that are already there.
Here a generalization.

Let $n$ be a positive integer and let $(r_1,\dots,r_n)$ be an $n$-tuple of nonnegative integers with:$$n=\sum_{k=1}^n kr_k$$
Then the number of permutations in $S_n$ that have cycle structure $(r_1,\dots,r_n)$ in the sense that their complete factorization into disjoint cycles contains $r_k$ cycles of length $k$ equals:$$\frac{n!}{\prod_{k=1}^nk^{r_k}\times r_k!}$$
An element in $S_{10}$ with order $10$ has one of the following structures

*

*$(0,0,0,0,0,0,0,0,0,1)$

*$(3,1,0,0,1,0,0,0,0,0)$

*$(1,2,0,0,1,0,0,0,0,0)$
Substitution gives us the following corresponding numbers:

*

*$\frac{10!}{10}=362880$

*$\frac{10!}{(1^3\times3!)(2^1\times1!)(5^1\times1!)}=\frac{10!}{60}=60480$

*$\frac{10!}{(1^1\times1!)(2^2\times2!)(5^1\times1!)}=\frac{10!}{40}=90720$
So in total there are: $$514080\text{ elements in }S_{10}\text{ that have order }10$$
A: Idea is that disjoint cycles commute. 
Case 1: $a=10,b=1$
We have only $10!$ such permutations. But since $(1,2,3,...,10)=(2,3,...,10,1)=...(10,1,2,...,9)$,i.e.each of the permutations repeats 10 times and hence total no. of order 10 permutations = $10!/10=9!$
Two more possibilities: Case 2(i): We have one disjoint cycle of order $5$, 2 disjoint cycles of order 2 each and one identity cycle of order $1$
We have $10!/(5!5!)\times5!\times 5!/(2!3!)\times 2!\times 3!/(2!1!)\times 2!=10!$ 
Each permutation for $5$ order cycle repeats $5$ times and that for $2$ order cycle twice and hence we have total no. of order 10 permutations = $10!/(5\times 2\times 2\times 2)=9!/4$ as two disjoint cycles of order 2 commute with each other. 
Case 2(ii): One cycle of order $5$ and one cycle of order $2$ and rest are identity. 
We have total no of possible permutations of order 10 =$^{10} C_5 \times ^5C_2 \times 5!\times 2!=
10!/5!5!\times 5!\times 5!/3!2!\times 2!=10!/3!$ and considering the repetitions we have $^{10} C_5 \times ^5C_2 \times 5!/5 \times 2!/2=9!/3!
$
So total is $9!+9!/4+9!/3! $
