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There will be more than one safe sequence (4 to be precise), try with banker's algorithm..

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After P3 more than one option possible as P2,P1,P4 all can be satisfied so many combinations possible hence the answer. Hope it helps!

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D) more than 2 safe sequence.

First P3 shall complete and then other 3 processes in any order.

safe sequences:

(P3 P1 P2 P4), (P3 P1 P4 P2)

(P3 P2 P4 P1), (P3 P2 P1 P4)

(P3 P4 P2 P1), (P3 P4 P1, P2)
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(D)More than 2 safe sequences exist.

Explanation: After creating the Request and allocated table from the request allocated graph given above we can see R1, R2 and R3 are left with only 0,1 and 0 resources respectively. With this, we only can satisfy the need for P3. After P3 executes it releases 1 instance of R2 and one instance of R3 making the available resources of R1, R2, and R3 as 0,2 and 1 respectively. With this, we can satisfy the requirement of all the processes so we can schedule any one of them so there will be more than two safe sequences possible.
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Yess more than 1 safe sequence possible.

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