How many DFA's exits with two states over input alphabet $\left \{ 0,1 \right \}$
For n states and m input alphabets we can have the formula:
$n*n^{nm}*2^{n}$
=$n^{mn+1}*2^{n}$
In a $DFA$ there might not be a difference if start state changes- as states are unlabeled usually. In such a case, we can divide the above by number of states giving $ n^{mn}×2^{n}$ possible $DFAs$.
Given $n=2,m=2$
$ n^{mn}×2^{n}$
$ 2^{2*2}×2^{2}$
$16*4$
=$64$
Option D
https://gateoverflow.in/10853/how-many-dfas-exist-with-three-states-over-the-input-alphabet
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