$2$ -input XOR $\rightarrow F=A'B+AB'=m_1+m_2 \text{ [2 out of 4 min terms]}$
$3$ -input XOR $\rightarrow F=A \oplus B \oplus C = \Sigma m (1, 2, 4, 7) \text{ [4 out of 8 min terms]}$
(i.e.,) any EX-OR , EX-NOR gate contains half of total number minterms $6$ input EX-OR is $\frac{2^6}{2}=32 \text{ out of 64 minterms}$
Formula:
N-input EX-OR Gate contains $\frac{2^n}{2}$ number of minterms at its output.
