Can we calculate the total number (i.e maximum no.) of dfa's that can be designed when the following constraints are put: |Q|=2{No. of states is 2},|Ɛ|=2{No. of alphabets} and |F|=1{No. of Final States} ?
First of all, I am guessing that by "No. of alphabets," you are actually referring to the number of symbols in THE alphabet. I haven't heard of a finite automaton with multiple alphabets.
Next, the definition of a finite automaton that I have is: A finite automaton M is a quintuple M = (S,I,δ,s0,F) where: S is a finite set (of states) I is a finite alphabet (of input symbols) δ: S × I → S (next state function) s0 ∈ S (the starting state) F ⊆ S (the accepting states).
So your definition maps to mine in that Q -> S Ɛ -> I and F -> F
Now, which state is the starting state leads to different automata, so this is an important factor and cannot be left out. If you have 2 states, then choosing a different final state out of the two leads to two different automata. Now assuming that there must be a transition function for each symbol in the alphabet for each state, then examining just a single state to begin with, for each state, each of the two symbols (call them a and b) must have a transition function. The value of the transition function for each symbol can be one of two possible states. Therefore, for a single state, there are a possibility of 2 x 2 = 4 transition functions. Since there are two states, there is another 4 possible transition functions for the second state. Accounting for the possibility of different initial/final states, there are 8 x 2 x 2 = 32 possible DFAs that you may design.