Ans:(b)
Explanation: The DFA's are defined by the set of final states and the transitions. (Assuming there is a designated starting state).
We know transition of DFA's are of the form $\delta(q_i,a)=q_j$. Now given there are 10 states and 5 symbols. $\therefore |Q\times \Sigma|=5\times 10=50$. Now each of these two-tuples can map to any of the 10 states . Therefore there are $10^{50}$ possibilities.
Now the set of final states can be a subset of the 10 states. Therefore total possibilities=$2^{10}$.
Hence, the required answer is $2^{10}\times10^{50}$