Some one ask me to count the number of element in the set \{\sigma\in S_5:|\sigma|=2\}, after doing that I follow the same idea to count |\{\sigma\in S_n:|\sigma|=2\}|, and a step further I count |\{\sigma\in S_n:|\sigma|=p\}|, p\leq n, and I get \sum_{k=1}^{[\frac{n}{p}]}\frac{\big((p-1)!\big)^{k}}{(k)!}\prod_{r=0}^{k-1}\binom{n-pr}{p}.
The result is summarized in the following document (I hope the result is true):
http://ihome.ust.hk/~cclee/document/sthgen.pdf
To ``verify" the result, let's count L(n,3) in another way, we count this by considering a_n defined by
a_n=\frac{1}{3}\times \underbrace{\binom{n}{2}}_{\text{choose 2 elements}\atop \text{to form permutation}}\underbrace{\binom{2}{1}}_{\text{choose 1 of the first two}\atop\text{ chosen numbers}}\underbrace{\binom{n-2}{1}}_{\text{choose another 1 to form permutation}\atop\text{with the number in 2C1 }}.
The factor \frac{1}{3} is left there becasue we observe that every length 3 permutation can be written as product of 2 transpositions in exactly 3 ways.
Then clearly the number of ways to form order 3 permutation by multiplying k disjoint length 3 cycles is given by \frac{\prod_{r=0}^{k-1}a_{n-3r}}{(k)!}=\frac{2^{k}}{(k)!}\prod_{r=0}^{k-1}\binom{n-3r}{3},
exactly the same summand appear.
