$Answer$ : $105$
This problem is same as Number of ways of partitioning the $8$ vertices into $4$ sets of two vertices each.
Now, vertex $1$ can do partitioning with any of the $7$ vertices (say vertex $1$ and $2$). So, now our base set is reduced by $2$.
Similarly, vertex $3$ can do partitioning with any of the remaining $5$ vertices. and vertex $5$ can do partitioning with remaining $3$ vertex.
So, number of ways of partitioning $=7\times5\times3=105$.