A subgroup of $S_6$ must, of course, contain the identity permutation id.
A subgroup of order three, say $A\leq S_6$, must must then contain two additional permutations, $\varphi$ and $\tau$ such that $\varphi^3 = \tau^3 = id$, each of which generates $A$, and such that $\varphi$ and $\tau$ are inverses of each other. Take, for example, the two 3-cycles $\varphi = (1 , 2, 3)\in S_6$ and $\tau = \varphi^2 = (1, 3, 2) \in S_6$. They are inverses of each other; you can check $\varphi ^3 = \tau ^3 = id.$ So each is of order $3$, and $\varphi \circ \tau = (1, 2, 3)(1, 3, 2) = \tau \circ \varphi = id $, so they are inverses of each other. Hence $A = \{id, (1, 2, 3), (1, 3, 2)\}$ is a subgroup of $S_6$.
Hence, $A$ is a subgroup of $S_6$ of order $3$. Your task is to determine how many such subgroups of order $3$ exist in $S_6$.
Theorem: The order of a cycle is equal to its length. The order of a product of disjoint cycles of lengths $i, j$ is equal to the $\text{lcm}(i, j)$.
So in $S_6$, (a) all permutations that are cycles of length $3$, and (b) all permutations that are the product of two disjoint cycles each of length $3$, like, say, $(2, 4, 6)(1, 3, 5)$, are of order $3$, and no other cycle-types are. Every subgroup of order three contains two distinct 3-cycles and the identity, or else two distinct products of disjoint 3-cycles.
Count all such permutations of type (a) or (b) in $S_6$ divide by 2, and you'll have the number of subgroups of $S_6$ whose order is $3$.
Think like that-what are the possible cycle structures of the permutations in $S_4$? There are only five possible options:
Four cycles of length $1$
One cycle of length $2$ and two cycles of length $1$
Two cycles of length $2$
One cycle of length $3$ and one cycle of length $1$
One cycle of length $4$
As you know, the order of a permutation equals to the lcm of the lengths of its disjoint cycles. So it is easy to check only the elements of the fifth type (one cycle of length $4$) are elements of order $4$. So all you need to find is how many cycles of length $4$ there are. Well, if you want $\sigma$ to be a $4$-cycle in $S_4$ you have $3$ options to choose the value of $\sigma(1)$, then $2$ options to choose what will be $\sigma(\sigma(1))$, and that's it. Once you know what are $\sigma(1)$ and $\sigma(\sigma(1))$ you know the whole permutation because it is a $4$-cycle and $\sigma(\sigma(\sigma(1)))$ must be the remaining element of $\{1,2,3,4\}$. So the number of such permutations is $3\times 2=6$.
Best Answer
The possible elements of $S_5$ are given by the various partitions of $5$. For example, $(i_1i_2)(i_3i_4i_5)$ is an element of $S_5$ corresponding to the partition $5=2+3$.
To count the number of such elements, we have $5$ choices for $i_1$ after which we have $4$ choices for $i_2$. We must then divide by $2$ because $(i_1i_2) = (i_2i_1)$. Hence we have $(5)(4)/2 = 10$ possibilities for $(i_1i_2)$. Once we have chosen $i_1$ and $i_2$, we then have $3$ choices for $i_3$, 2 choices for $i_4$ and one remaining choice for $i_5$ giving $6$ possibilities for $(i_3i_4i_5)$. However we observe that, for example, $(345) = (534) = (453)$ (i.e. each three cycle yields two other equivalent 3-cycles). Thus we have $6/3 = 2$ possibilities for $(i_3i_4i_5)$.
This gives $$\frac{(5)(4)}{2}\cdot 2 = 20$$ possibilities for elements of the form $(i_1i_2)(i_3i_4i_5)$.
Can you generalize this for other elements of the group?