Bankers Algorithm

Question

What is the total safe sequence possible?

 

Comments

Total resources available before allocation is (10,10,10)

Remaing resources available (10-6,10-6,10-5)=(4,4,5)

So, you can't start with p2.

If you starts with p1 or p3 or p5, then second one can't be p2.

P2 should be after p4.

With that observation, it make $\frac{5!}{2} = 60$

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Why not P2, P2 requires only less than available

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Yes, you are correct.

I mis-read the table. The given table have maximum demand. But I thought it is remaining requirements.

Then 5! is correct.