I will use theorem that the order of permutation of a finite set written in disjoint cycles is the least common multiple of the lengths of the cycles.
I will use notation $(n)$ to represent a cycle of length n. So we can decompose some permutations of $S_{10}$ as a product of disjoint cycles such that the least common multiple of their lengths is 10, and sum of lengths of each disjoint cycle is 10.
So, there are 3 possibilities:
1)$(5)(2)(2)(1)$,
2)$(5)(2)(1)(1)(1)$,
3)$(10)$.
- ${10 \choose 5} { 5 \choose 2} { 3 \choose 2} \frac{1}{2} $ permutations here. I divide by 2, because there are two disjoint cycles of length 2, and they are repeating themselves 2 times, so we exclude these cases.
- $ {10 \choose 5} {5 \choose 2} $
- ${10 \choose 10} = 1$
then we sum 1. 2. and 3. . But I think my logic somewhere is terribly wrong could you please point out where exactly? I am not sure that this is the right way to calculate number of such permutations.
We know that $\sigma, \theta\in S_{10} $ such that $\sigma $ and $\theta $ are disjoint permutations then $\mathcal{O}(\sigma \theta)=LCM(|\sigma|,|\theta|)$ Finding all possible $\textbf{LCM} =10$ We have $|\sigma|=5$ and $|\theta|=2$ number of these type of elements are $\displaystyle\frac{10\times9\times8\times7\times6}{5} \times\frac{5\times 4}{2}=\frac{9!}{3!}$ and other type of cycle $|\sigma|=10$ and $|\theta|=1$ $9!$ Next possibility $|\sigma|=5$ , $|\theta|=2$ and $|\gamma|=2$ Here number of elements are $\frac{10!}{2!\times 5}=\frac{9!}{2}$ Adding these possibilities we will get our desired result