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There are 3 final states in total 10 states .

7 are non final states , so all these state which come together to form a state in dfa are not have any finale state in dfa. .

so possible DFA state with only these 7 Non final states are 2^{7}^{ }= 128

and total 2^{10 } = 1024 states are possible

so total number of state which contain final state is 1024 - 128 = 896

option C is true here.

@Abhisek Tiwari 4; please note that the states are provide for NFA.

When we convert NFA to DFA; if there is a transition in NFA which goes to Φ or empty; then in corresponding DFA it is a reject state.

Thus the subset can consist of a Φ state.

I attempted this question in a bit longer method:

Number of final state is 3 (8,9,10)

Number of non-final state is 7

Since, Accepted state consist of one or more combination of 8,9,10 (2^3 combinations - 1 {removing the Φ state as then there will be no combination of final state and this will not be final sate in DFA}) and 0 or more combinations of 1 to 7 (2^7): total = 2^3 * (2^7-1) = 7*128 = 896

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