Looking at the precedence Graph, we get
So from here, we can see that any transaction of T1 should happen before T2 & T3 and any transaction of T2 should happen after T1 and before T3 and any transaction of T3 should happen after T1 & T2.
Let us ignore W4(C) for now since that can occur anywhere among 7 possible places.
So for | W1(A),W1(B) | R2(A),W2(B) | R3(A),W3(B) | none of the separated Read/Write can occur in/before/after the next/previous separated section since that will violate our above condition.
So possible permutations are limited only among the Reads/Writes of One single Underlined section.
So Possible number of permutations are 2*2*2=8
Consider W4(C) and its 7 possible places, we get 7*8=56 possible permutations.