first don't change anything. then R4(A),R2(A),R3(A),W1(B) can be arrange in 4! ways;
now if we shift W2(A) before W1(B) no change in conflict serializability.but now we have onle 3 to permute therefore 3!
if we shift W2(A) with R3(B) again no change in conflict serializability . now we have 5 things for permutation. number of permutation are 5!/2! as R3(A) and R3(B) are from same transaction hence order is preserved.
total=24+6+60
90
if you find anything wrong plz do comment