Number of elements of order 2 in $ S_6$ Find number of Elements of order 2 in $ S_6$.
My attempt:
It is clear that permutations would be of form:


*

*(a b)

*(a b)(c d)

*(a b)(c d)(e f)
Option 1. Will have $\frac{6!}{2.4!}$ permutations
Option 2. Will have $\frac{ \frac{6!}{2.4!} . \frac{4!}{2.2!}}{2}$
Option 3. Will have $\frac{ \frac{6!}{2.4!} . \frac{4!}{2.2!} . \frac{2!}{2}}{3}$
So adding them total is 90. But answer in book is 75. What is going wrong?
 A: In the third case, there are $3!$ possible orderings of the three transpositions, so you should be dividing the quantity in the numerator by $3!$, not $3$.
In fact, you can save some computation by observing the following: Let $n_k$ be the number of distinct products of $k$ transpositions.
First, there is a bijective correspondence
$$\{(ab)(cd)(ef)\} \leftrightarrow \{(cd)(ef), (ab)(ef), (ab)(cd)\}$$
given by taking a product of three transpositions, and separately pulling out each of the three transpositions. Thus, $n_2 = 3 n_3$.
Likewise, there is a bijective correspondence
$$\{(ab)\} \leftrightarrow \{(cd)(ef), (ce)(df), (cf)(de)\}$$
given by taking a transposition $\lambda$ and writing out the three products of the transpositions built out of the four elements not in $\lambda$. So, again, $n_2 = 3 n_1$ and thus $n_1$ = $n_3$.
So, the total number of elements of order $2$ is
$$n_1 + n_2 + n_3 = n_1 + 3 n_1 + n_1 = 5 n_1,$$ but $n_1 = 15$, so the total number of elements is $$5 (15) = 75$$ as desired.
