How many elements of order $2$ does Sym $6$ have? First, I will answer the following question:
''How many elements of order $2$ does Sym $5$ have?''
The answer is:
$(12),(13),(14),(15),(23),(24),(25),(34),(35),(45),(12)(34),(12)(35),(12)(45),(13)(24),(13)(25),(13)(45),(14)(23),(14)(25),(14)(35),(15)(23),(15)(24),(15)(35), $
that is, there are 22 elements of order $2$ does Sym $5$ have.
I omitted 3 products of two transpositions; the correct number for S5 is 25. Thanks @BrianM.Scott


*

*How many elements of order $2$ does Sym $6$ have?
I can compute as a manual but it will be too long. Is there any easy method to find it? Thanks...


One can always count all the elements of order $2$ in $S_6$.
Since we have six elements to play with, it's clear that elements of order two look either like $(ab)$, $(ab)(cd)$ or $(ab)(cd)(ef)$.

*

*First we find all the elements that look like $(ab)(cd)(ef)$. There are ${6\choose 2} = 15$ ways to create a 2-cycle $(ab)$. To get $(ab)(cd)$ we have ${4\choose 2}=6$ options left. Next, there's only one possible choice left to get $(ab)(cd)(ef)$. Finally, by canceling the $3!$ repetitions due to ordering 3 cycles, we get
$$
\frac{15 \times 6 \times 1}{3!} = 15
$$


*Similarly, for $(ab)(cd)$ we have
$$
\frac{15 \times 6}{2!} = 45
$$


*Finally, for $(ab)$ we have
$$
{6\choose 2} = 15
$$
Therefore there are $15 + 45 + 15=75$ elements of order $2$ in $S_6$.
