This exercise asks to provide elements of order 10, 20, and 30 in $S_{10}$. Thinking that the order of a permutation $\sigma$ is the least common multiple of the length of the disjoint cycles whose product is $\sigma$, I went for the following:
$\sigma_1=(a_1a_2)(b_1b_2b_3b_4b_5)$ of order 10
$\sigma_2=(a_1a_2a_3a_4)(b_1b_2b_3b_4b_5)$ of order 20
$\sigma_2=(a_1a_2a_3)(b_1b_2b_3b_4b_5)(c_1c_2)$ of order 30
The question also asked if there could be a permutation of order 40. This is where I got stuck, mainly because I could not find any combination of cycle lengths which would give 40 as least common multiple. Because the order of $S_{10}=3628800$, and 40 divides that, I figured there would have to be an element of order 40, and much larger orders as well, which it seems to be impossible to obtain considering a permutation written as disjoint cycles. Help in finding the flaw in my reasoning would be appreciated.