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answer is 64,

every alphabet has two choices on every state . as on state q1 ,0 can either go to q0 or remain on q1, so at every state 4 choices are there, ans such 2 states are there so 4*4=16 is the total number of transition possible, now for final state we have 4 possibility, while for inital state we have just one possibility. so it will be 1*4*4*4=64. if no boundation on initial state it will be 128. (2*4*4*4)
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for a state, on an input symbol we have two choices, which are

  1. end up in the same state
  2. end up in the other state

There are two input symbols available.
So, total number of ways transition is possible from a state = $\text{Choices for an input symbol} \times \text{Number of Input Symbols} = 2\times 2 = 4$

Total number of ways in which transition can happen in a DFA with two states = $\text{Choices at state }q_0 \times \text{Choices at state }q_1 = 4 \times 4 =16$ 

In a DFA any state can be a final state or an initial state. Here, Input state is fixed, so we need not worry about it.
total number of possibilities for the set of final states = $4$

Hence, total states in this DFA = $4 \times 16 = 64$
 

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