Let $M$ be a nondeterministic finite automaton (NFA) with 6 states over a finite alphabet. Which of the following options CANNOT be the number of states in the minimal deterministic finite automaton (DFA) that is equivalent to $M$ ?

For any NFA $M$ with $n$ states, the powerset construction (subset construction) produces an equivalent DFA with at most $2^n$ states.

Given $n = 6$ states:

Since the minimal DFA equivalent to $M$ cannot have more states than the DFA obtained via subset construction, the number of states $k$ in the minimal DFA must satisfy:

Evaluating the given options for what CANNOT be the number of states:

• (A) 32: Possible, since $32 \le 64$.

• (B) 65: Impossible, since $65 > 64$.

• (C) 1: Possible (e.g., if $L(M) = \Sigma^*$ or $L(M) = \emptyset$), since $1 \le 64$.

• (D) 128: Impossible, since $128 > 64$.

Therefore, the options that CANNOT be the number of states are B and D.