How many different finite automata are there that have 3 states $q_0$, $q_1$ and $q_3$, have an alphabet of size 2, and where $q_0$ is starting state?
Total number of distinct transition functions from $Q \times \Sigma \rightarrow Q$ is $3^6 = 729$. This is because $Q \times \Sigma $ has $|Q| \times |\Sigma| = 3 \times 2 = 6$ elements, and $|Q| = 3$. Every one of $6$ elements has $3$ choice for transition.
Even if there are 3 different choice of start state, that will not give us different automata, as these automata are already counted in 729 different automata with renumbering of states. So let us assume $q_0$ is the start state. Now $q_0$ is always reachable. We will remove those automata where some of the states are reachable (because we can compute these numbers easily, that is why, otherwise if we are not interested in that, we can straight away have $3^6 \times 2^3 = 5832$ distinct automata).
Number of distinct transition functions where 2 states are unreachable = $1^2.3^4 = 81$
Number of distinct transition functions where $q_1$ state is unreachable = $2^4.3^2 = 144$
Number of distinct transition functions where $q_2$ state is unreachable = $2^4.3^2 = 144$
Number of distinct transition functions where every state is reachable = $729 - 2\times144 + 81 = 522$.
Now let us count different final states.
3 final states = $522$ different automata (though language is same, however we are counting different automata, language being same is irrelevant).
1 final state = $522$ ($q_0$ final) + 522 ($q_1$ final) = $1044$ ($q_2$ final state is counted in $q_1$ final state, as it will just be renaming of state $q_1$).
2 final states = $522$ ($q_0$ and $q_1$ final states) + $522$ ($q_1$ and $q_2$ final states) = $1044$ (again we won't count $q_0$ and $q_2$ as final states).
0 final states = $522$ (perfectly valid different automate, and again languages are all same).
So total number of distinct automata is $522 + 1044 + 1044 + 522 = 3132$.
As an added note, the case of all possible two state automata is also interesting, with any alphabet size. The interesting two cases of FA will be as following:
In the automata above, $S$ is the set of all "Set" transition symbols, $R$ is the set of all "Reset" transition symbols, $T$ is the set of all "Toggle" transition symbols, and $U$ is the set of all "Unchanged" transition symbols. We set two automata depending on whether we want set condition or reset condition to be the final state. The corresponding regular expressions are
$(R+U)^*(S+T)((S+U)+(R+T)(R+U)^*(S+T))^*$ for first automata, and,
$((R+U)+(S+T)(S+U)^*(R+T))^*$ for second automata.
It is difficult to do such neat analysis for 3-state automata mainly because there will be 3^3=27 different classes of transition symbols.