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?
 A: Let the $2k$ elements of $X$ be people. We want to count the number of ways to divide these $2k$ people into $k$ teams of $2$ each.
Line up the people from left to right in order of age, or weight, or student number.
The leftmost person can choose her team mate in $2k-1$ ways.
For every such way, the leftmost person not yet chosen can choose her team mate in $2k-3$ ways.
For every choice made so far, the leftmost person not yet chosen can choose her team mate in $2k-5$ ways.
And so on.
If the "and so on" is not viewed as rigorous enough, we rewrite the proof as a proof by induction.  We assume that the result holds for $n=2k$, and prove that the result holds for $n=2(k+1)$. Alicia  has $2k+1$ choices of team mate. For every choice, there are $2k$ people left, and by the induction assumption they can be divided into teams in $(2k-1)(2k-3)\cdot (3)(1)$ ways.  Thus the $2(k+1)$ people can be divided into teams in $(2k+1)(2k-1)\cdots(3)(1)$ ways. This completes the induction step.
Remark: In your example, we can list the divisions into teams explicitly. Let the people be $A, B, C, D$. Then $A$ can be paired with $B$, $C$, or $D$. Once this is done the other pairing is determined.
A: 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.
