No. of conflict serializable = total schedule - no. of non conflict schedule
Total schedule = (m+n)!/m!n! = (4+4)!/4!4! = 70 ,here m and n is no. of operations in t1 and t2
now no. of non conflict schedule can be calculated as:
T1:
pos1
R1(X)
pos2
W1(X)
pos3
R1(Y)
pos 4
W1(Y)
pos 5
For non conflict pair,R2(Y) can place at pos 1 or 2 or 3 or 4 and for W2(Y) ,it must place at either pos(4) or pos(5)
means for one position of R2(Y) there is two choices for W2(Y) ,then total combination for non conflict pair possible = 2 * 2 *2*2=16
then total conflict shedule = 70 - 16 =54. Answer