Total View Serializable schedules will be 6. (Refer: https://gateoverflow.in/13348/solve)
These 6 are:
T1->T2 T3->T4 T5->T6 1st order
T1->T2 T5->T6 T3->T4 2nd order
T5->T6 T3->T4 T1->T2 3rd order
T3->T4 T5->T6 T1->T2 4th order
T5->T6 T1->T2 T3->T4 5th order
T3->T4 T1->T2 T5->T6 6th order
Out of these, except for the 1st order, none of the others are conflict equivalent to the given schedule. This is because the given schedule has conflicting Read-Write operations successively and hence it can't be transformed into another conflict equivalent schedule via swapping.
Therefore, answer should be 5.