In a group of $9$ people, each person shakes hands with exactly $2$ of the other people from the group. Let $X$ be the number of possible ways to perform these handshakes. Take $2$ handshake patterns (arrangements) distinct if and only if at minimum $2$ people who shake hands under one pattern (arrangement) don't shake hands under the other pattern (arrangement). Find $X$.
I think casework is the way to go.
$A$ shakes with $B$ & $C$. $D$ shakes with $E$ & $F$. $G$ shakes with $H$ & $I$.
Perhaps I could use a recurrence relation, but I don't see a possible way.
In total there are:
$$\binom{9}{3} \cdot \binom{6}{3} \cdot \binom{3}{3} = 1680$$
Ways to choose groups of three people.
But I dont anything else to this problem, and clearly this is the wrong answer.
Hints only please!