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there are totally 'n' states, 'm' alphabets.

  • Each state can be final or non-final state (2 choices). So, there are 2^n variations.
  • for each state, there is 'm' out edges (alphabets) having 'n' destination nodes to choose from. So, each state will add 'n * m' variations which gives us n^n*m variations.

In total, we will have 2^n * n^n*m DFA's.

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