How many disjoint subsets? I have a question about combinatorics.
I have the following set:
$$M = \{ 1,2,...,n \}$$
How many disjoint subsets
$$A\subseteq M, \quad |A| = 2$$ 
are there? For the future, how do I approach questions like these?
Thanks in advance for your help 
EDIT:
$n$ is even. I want to know how many ways there are to split $M$ into disjoint subsets of size $2$. If this is any help, I am trying to figure out how many permutations in $S_n$ exist, that are involutions as well as derangements. For $n$ odd there are none, that's what I've shown so far. For $n$ even, I can write the permutation in brackets of size two (I don't know how you call this way of writing a permutation down).
The order of the sets do not matter.
 A: Let $A_n$ be the number of ways of partitioning an $n$-element set $M$ into $2$-element subsets. Pick $x \in M$. There are $n - 1$ choices for the element that is going to belong to the same partition as $x$, and then the remaining $n - 2$ elements can be partitioned into $2$-element subsets in $A_{n - 2}$ ways. Clearly $A_1 = 0$ and $A_2 = 1$, so we have:
\begin{align}
A_n &= (n - 1)A_{n-2}\\
A_2 &= 1\\
A_1 &= 0
\end{align}
Hence $A_n = 0$ for odd $n$ and for even $n$ we have:
$$
A_{n} = 1 \cdot 3 \cdot \ldots \cdot n -1.
$$
A: Since $n$ is even, say $n=2k$ for some integer $k$.  How many ways are there to pick a pair of these integers?
$$
\binom{2k}{2}
$$
How many ways are there to pick a pair from the remaining numbers?
$$
\binom{2k-2}{2}
$$
So, the number of tuples (order matters for now) of the form $(A_1, \dots, A_k)$, where each $\lvert A_i \rvert = 2$ is given by the multinomial coefficient
$$
\binom{2k}{\underbrace{2 \; 2 \; \cdots \; 2}_{k}} = \binom{2k}{2} \binom{2k-2}{2} \cdots \binom{2}{2} = \frac{(2k)!}{2! \; 2! \; \cdots \; 2!}.
$$
To get the number of sets of the form $\{ A_1, \dots, A_k \}$, where each $\lvert A_i \rvert = 2$, you measure the redundancy.  How many times did each size $k$ set get counted as a $k$-tuple?  The number of permutations of size $k$ is $k!$, so the number of involutive derangements is
$$
\frac{1}{k!} \binom{2k}{\underbrace{2 \; 2 \; \cdots \; 2}_{k}} = \frac{(2k)!}{k! \; (2!)^k}.
$$
Miraculously, this expression simplifies further!  Notice that the denominator is equivalent to the product of the first $k$ even numbers $2 \cdot 4 \cdots 2k$; hence your count is is just the product of the first $k$ odd numbers
$$
(2k - 1)!! = 1 \cdot 3 \cdots (2k - 1).
$$
Here are the first couple values in this sequence:
\begin{array}{c|rrrrr}
k & 1 & 2 & 3 & 4 & 5 \\
\hline
(2k-1)!! & 1 & 3 & 15 & 105 & 945
\end{array}
