How many of subgroups of order $2$ are there in $S_4$? Any subgroup of order $2$ will be cyclic subgroup and so will be generated by single element of order $2$ in $S_4$, so to count number of subgroups of order $2$ we need to count number of elements of order $2$ in $S_4$, I tried counting them but I got answer $8$ but is $9$ actually.
 A: Assume that $\sigma\in S_4$ has order two. Then $\sigma$ changes some element $x$. Let $y=\sigma(x)$. Then $\sigma(y)=x$.
This shows that $\sigma$ is the product of some disjoint $2$-cycles.
The possibilities are:
$$(1,2),(1,3),(1,4),(2,3),(2,4),(3,4),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)$$
A: Your starting point sounds good; we only need to consider elements of order 2. If we consider a sequence of four numbers, these elements can be thought of as the "swapping" permutations.  
There are $ {4 \choose 2} = 6$ ways to choose a pair of numbers which swap and leave the other values alone. Then for each of these, we can also consider swapping the remaining two values. This gives another 6/2=3 (dividing by two to prevent double counting) permutations of order 2. Giving a total of 9.
A: The permutations (elements) of order two in $S_4$ are the following:
$$\underbrace{(12), (13), (14), (23), (2 4), (34)}_{6 \;\;\text{two-cycles}}, \underbrace{(12)(34), (13)(24), (14)(23)}_{3\;\; \text{products of disjoint two-cycles.}}$$
As Fransesco mentions in a comment above: We see  "six transpositions + three products of two disjoint transpositions."
