Elements that are their own inverses in a symmetric group. How many elements are their own inverses in $S_6$? I'm having a hard time figuring out how to calculate such a thing.
Here are the key facts:


A permutation is its own inverse iff it has order $2$. (Such permutations are called involutions.)
The order of a permutation is the lcm of the orders of the cycles in its cycle decomposition.


This should give you the type of all possible cycle decompositions.
In S6, here are three types of permutations of order 2: a the transpostions (ij) that swap the digits i and j (for i and j different digits between 1 and 6), the double transpostions (ij)(kl), where i, j, k, l are four different digits between 1 and 6, and the triple transposition (ij)(kl)(mn), where i,j,k,l,m,n are six different digits between 1 and 6.

Obviously the number of transpositions (ij) is 6 . 5 / 2, which is 15.

For the double transpositions, the transformation of order 2 will be $(1/2((6.5)/2)((4.3)/2))$ =45.

For the triple transposition, it will be 
$(1/3((6.5)/2)((4.3)/2)((2.1)/2))=30$

And also since identity is self inverse element too 
Therefore total number of elements of order 2 in S6 = 15+45+30+1 =91.