How to count the number of products of disjoint transpositions in $S_5$?

Question

How to count the number of elements of $S_5$ having the form $(i j)(k l)$ where $i,j,k$ and $l$ are all distinct?

Context: I have to list representatives of the conjugacy classes in $S_5$ along with the cardinality of each one. I had success with representatives of $5$, $4$, $3$-cycles, $3$- cycles times transpositions and transpositions (if I'm not mistaken, they are $24$, $30$, $20$, $20$ and $10$, respectively). However I could not do the above one.

I know that the answer is $15$ (by subtracting $105$ from $120=5!=|S_n|$), but I'd like to know how to count them.

Thanks for help

EDIT: I think I solved it. There are ${5 \choose 2} \cdot {3 \choose 2}$ numbers to pick, and to avoid repetition (e.g. we have $(12)(34) = (34)(12)$, etc.) we divide the resultant by $2$ to get $15$.

Answer 1

Here's another method that generalizes well. Let $S_5$ act on $1,2,3,4,5$ in the usual way. Then the stabilizer of $(12)(34)$ is generated by $(12)$, $(34)$ and $(13)(24)$. That is, we can switch $1 \leftrightarrow 2$ and $3 \leftrightarrow 4$ and interchange the two cycles without any effect. Thus the stabilizer has size $8$. Hence the orbit has size $\require{cancel}5!/8 = 5 \cdot \bcancel{4} \cdot 3 \cdot \bcancel{2} \cdot 1 = 15$.

Answer 2

There are the same number of these permutations as there are of 3-cycles, because you can pair, for example, $(123)$ with $(123)(45)$.

And there are 20 3-cycles, because there are ${5 \choose 3} = 10$ ways to pick the elements in the cycle and then 2 ways to make the cycle.