S$_{1}$ : R$_{1}$(X), W$_{1}$(X), R$_{1}$(Y), R$_{2}$(Y), W$_{1}$(Y), W$_{2}$(Y)

S$_{2}$ : R$_{1}$(X), W$_{1}$(X), R$_{2}$(Y), R$_{1}$(Y), W$_{1}$(Y), W$_{2}$(Y)

these two schedules are conflict equivalent.

in other words, i can say, S$_{2}$ is conflict equivalent to S$_{1}$

number of schedules which are conflict equal to S$_{1}$

if you follow the approach like in the following link, you will found answer as 4

https://gateoverflow.in/118640/gate2017-2-44?show=289691#a289691

one is the original one, i.e., S$_{1}$ and one more is S$_{2}$

remaining two are

S$_{3}$ : R$_{1}$(X), R$_{2}$(Y), W$_{1}$(X), R$_{1}$(Y), W$_{1}$(Y), W$_{2}$(Y)

S$_{4}$ : R$_{2}$(Y), R$_{1}$(X), W$_{1}$(X), R$_{2}$(Y), R$_{1}$(Y), W$_{1}$(Y), W$_{2}$(Y)

Note :- If a Schedule is not conflict serializable then NONE of it's conflict equivalent schedules are conflict serializable.

number of serial schedules which are conflict equal to S$_{1}$?

Here, you are Comparing Serial schedules with given Schedule like

for(i=0;i<n;i++)
{
if( a[i] == j)
{
}
else
{
}
}

here i is variying but not j. So, i represent Serial Schedule and j represent given Schedule S$_{1}$

There are 2 Transactions, So you have 2 ! = 2 Serial schedules. And None of them is equivalent to S$_{1}$

Let T$_{1}$ : R$_{1}$(X), W$_{1}$(X), R$_{1}$(Y), W$_{1}$(Y) and T$_{2}$ : R$_{2}$(Y), W$_{2}$(Y)

No.of Total Schedules Possible = $\frac{(4+2)!}{4!.2!} = \frac{720}{48} = 15 $

No.of Serial Schedules Possible = 2 ! = 2

Non-Serial Schedules = (10-2)=13 Schedules, in that Some are conflict equivalent to T$_{1}$ to T$_{2}$, Some are conflict equivalent to T$_{2}$ to T$_{1}$ and some of them not equivalent to any serial schedule.

let S : R$_{1}$(X), W$_{1}$(X), R$_{2}$(Y), W$_{2}$(Y), R$_{1}$(Y), W$_{1}$(Y)

number of schedules which are conflict equal to S

Answer : 6 including S

number of serial schedules which are conflict equal to S?

Answer is 1. ( T$_{2}$ to T$_{1}$ )

The total number of conflict serializable schedules that can be formed by T1 and T2 is

No.of conflict serializable schedules = No.of conflict equivalent to Schedule T$_{1}$ to T$_{2}$ + No.of conflict equivalent to Schedule T$_{2}$ to T$_{1}$ = 1 + 6 = 7

Non-Conflict serializable schedules = 15-7 = 8

Those 8 schedules are :

S$_{1}$ : R$_{1}$(X), W$_{1}$(X), R$_{1}$(Y), R$_{2}$(Y), W$_{1}$(Y), W$_{2}$(Y)

and it's remaining 3 conflict equivalent schedules.

S$_{5}$ : R$_{1}$(X), W$_{1}$(X), R$_{1}$(Y), R$_{2}$(Y), W$_{2}$(Y), W$_{1}$(Y)

and it's remaining 3 conflict equivalent schedules.