Rishi yadav
How 24 please explain ??
I think i am wrong kumar.dilip
I am getting 105
here's what I did :
fixed r(a) of T1, now r2(A) has three places - before r1(A), after r1(A) and after w1(B) because of two items r1(a) and w1(b)
now for r3(a), we have 5 places because of 4 items r1(a), r2(a) and w1(b) and w2(b)
similarly, for r4(a) we have 7 places because of 6 items
now total possible places would be - 3*5*7 = 105
Method 2 :-
Fo with Topological sort method,
Some more references to solve like this problems :-
https://gateoverflow.in/253496/number-of-topological-order
https://gateoverflow.in/245897/hashing-with-linear-probing
@Anuj Mishra
Can we start with $R_{3}$ after fixing $R_{1}$ and then $R_{2 }$ & then $R_{4}$ ??
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