Answer

Correct Option: (iii) $2^{n}$

Explanation

NFA with $n$ states → DFA via subset construction:

Each DFA state is a subset of NFA states. NFA states set $Q_N$, $|Q_N|=n$.

Number of distinct subsets = $|\mathcal{P}(Q_N)| = 2^{n}$.

This includes $\emptyset$ (dead state).

Hence DFA has at most $2^{n}$ states. Upper bound is tight: there exist NFA families (e.g., language “$n$th symbol from end is 1”) that require exactly $2^{n}$ DFA states.

Final Answer

Maximum $2^{n}$ states → Option (iii). (Minimal DFA may be smaller, but maximum is $2^{n}$.)