Double Factorial I am having trouble proving/understanding this question.

Let $n=2k$ be even, and $X$ a set of $n$ elements. Define a factor to be a partition of 
  $X$ into $k$ sets of size $2$. Show that the number of factors is equal to $1 \dot\ 3 \dot\ 5 \cdots\ (2k-1)$.

Lets suppose our set $X=\{x_1,x_2,...,x_n\}$ contains $4$ elements, i.e $X=\{x_1,x_2,x_3,x_4\}$ then $k=2$ implies $k = (x_1,x_2), \ (x_3,x_4)$ but the numbers of factors equals $3$. What am I doing wrong trying to understand this?

You can also partition the set as $\{x_1,x_3\}$ and its complement and as $\{x_1,x_4\}$ and its complement. Giving three factors in total. 
In general there are always $2k-1$ two element sets containing some fixed element. This observation also suggest a way towards a proof.
