This is equivalent to asking in how many ways a $2n-$element set can be partitioned into two-element subsets. Let $T_{2n}$ be that number.
Let's define
$T_0 = 1$
$T_2$ clearly equals $1$.
We can express $T_{2(n+1)}$ in terms of $T_{2n}$ and $T_{2(n-1)}$?
Let's consider the set with $2n+2$ elements $\{1,2,3,\dots,2n\}\cup\{a,b\}.$
If we choose $\{a,b\}$ to be in the partition, then we know that there are $T_{2n}$ ways to partition the remaining $2n$ elements.
We can choose partition elements of the form
$\{a,u\}$ and $\{b,v\}$ where $1 \le u \ne v \le 2n$
in $2n(2n-1)$ ways and there will always be $T_{2n-2}$ ways to partition the remaining $2n-2$ elements.
This implies that
$$T_{2(n+1)} = T_{2n} + 2n(2n-1)T_{2n-2}$$
The OEIS points out that
$$T_{2n} = (2n-1)!! = (2n-1)(2n-3)\cdots1$$
In either case
$T_{18} = 34459425$