I am trying to solve the following question regarding FSA:
"Suppose that $ S $ and $I $ are finite sets such that $|S| = n $ and $|I | = k$. How many different deterministic finite-state automata $M = (S, I, f, s0 , F )$ are there where the starting state $s0$"
I think the solution might be found using the product rule. The number of possible automata we can build is equal to the number of ways in which I can perform three different tasks: choosing the starting state $s0$, that is unique, chosing the set of final states $F$ and choosing the number of different functions from $S\times I$ to $S$ that we can find.
The number of all the possible $F$ is the cardinality of the power-set fo $S$: $2^{n}$
The number of all possible starting states is the cardinality of $S$: $n$
The number of possible functions from $S\times I$ to $S$ is $n$ multiplied with itself $ |S\times I| $ $ = nk$ times, so $n^{nk}$ is the number of possible functions.
Therefore, we should have $n^{nk} $ $ \times $ $n$ $ \times $ $2^{n}$ deterministic finite state automata.
Is this the correct solution to the problem?