Answer

Correct Option: (ii) Context-free but not regular

Explanation

$L = \{ a^m b^n c^{m+n} \mid m,n \ge 1 \}$.

Not Regular

Use Pumping Lemma. Let $p$ be pumping length, $w = a^p b^p c^{2p} \in L$. Pumping $y$ in $a$ segment breaks $m+n$ balance.

Context-Free

PDA construction:

1. Push $a$’s (m times) → stack $= A^m$.
2. Push $b$’s (n times) → stack $= A^m B^n$ (or same symbol).
3. For each $c$, pop one symbol. Accept if empty after $m+n$ $c$’s and no extra input.

Alternatively grammar:
$S \rightarrow a S c \mid a T c$
$T \rightarrow b T c \mid b c$

This generates $a^m b^n c^{m+n}$.

Not context-sensitive requirement beyond; CFL suffices, but not regular. Not Type-0/1 only.

Final Answer

Context-free but not regular → Option (ii).