self doubt(conflict serializable)

@shaik masthan i understood the second part T2->T1(53) of first question but i am not getting T1->T2 of first question why there is only one possibility is it because in same schedule they should follow the same order but my doubt is that r(a) w(a) and r(b)w(b) is there one on data item a and other on b so cant they permute beacuse they are on reading and writing on different data item so how it will make difference in answer 

why cant they should be like these R(B)W(B)R(A)W(A)R(C)W(C)R(B)W(B)

then we can get more possiblity 

 please tell me where i am going wrong

why there is only one possibility is it because in same schedule they should follow the same order

yes, with in a transaction you have to follow the order ===> after completing the operation R(B) it should go to W(B) then R(C) after that only W(C)

@Shaik Masthan

Can you please explain the solution to the 2nd question a bit?

I am not getting the cases..The way I solved is giving me more no. of schedules..

@Shaik Masthan

why you only take Write - Read conflict  here

By draw topological diag it's very easy to solve but I didn't understand that how you made it

acc to question 2

This topology is correct ???

but acc to this topology "correct answer is not coming :/ "

why you only take Write - Read conflict  here

that is DAG, So excess links removed !

if you didn't get it, just assume, 3 --> 6 you drawn a link.

Then think 6 should be after 3 and 4, and 4 should be after 3 ==> you must have to do 3--> 4--->6

Your DAG, representating T1 --> T2

@Shaik Masthan

In the first method, second image –

 https://drive.google.com/file/d/1Z-tlThOEyVlBjV8Zky5_Wc3z7L-hvSGc/view?usp=sharing

Total number of possibilities such that T2->T1 = T1->T2 = 56 is because of the symmetricity of the DAG, right? i.e if we draw the graph for T2->T1 it will be same as that of T1->T2 with the corresponding labels interchanged(G for A, H for B etc) , right?