TITLE: Involutions of $S_n$
QUESTION [7 upvotes]: How am I able to find how many involutions are in $S_n$ for any nonnegative integer $n$? Let's say for $S_4$. Is there an equation or anything that allows me to find this with ease?

REPLY [4 votes]: This is one of the first (if not the very first, I haven't got the book handy) examples in R.P. Stanley's Enumerative Combinatorics, Vol 1. There is no closed formula for the answer $a_n$ for a given value of$~n$, but better than to describe $a_n$ by a summation involving amongst others binomial coefficients, you have the power series expression
$$
   \sum_{n\in\Bbb N}a_n\frac{X^n}{n!}=\exp\Bigl(X+\frac{X^2}2\Bigr).
$$
See also this answer. There are fast methods to find the coefficients of such power series expression. In the current case they lead to the recurrence $a_n=(n-1)a_{n-2}+a_{n-1}$ and initial values $a_0=a_1=1$ that will rapidly give you the initial part of the sequence $(a_n)_{n\in\Bbb N}=(1,1,2,4,10,26,76,232,764,\ldots)$ Note that this requires fewer arithmetic operations, and simpler ones with smaller numbers, to compute all of $(a_0,\ldots,a_n)$, than the summation expression $a_n=\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}\frac{(2k)!}{k!2^k}$ requires to compute just$~a_n$.
The recurrence is also easily understood directly: if $n$ is a fixed point of an involution of $S_n$ then what remains is an involution of $S_{n-1}$, and otherwise $n$ is interchanged with one of the $n-1$ previous elements, and an involution of the remaining $n-2$ elements remains to be chosen.