option $a$ represents atmost one man
option $b$ represents exactly 2 men.
option $c$ represents atmost 2 men
Atmost 2 men $\left \{ 0 \text{man OR } 1 \text{men OR } 2 \text{men}\right \}$
$\forall x \forall y \forall z (\text{Male(x)} \wedge \text{Male(y)} \wedge \text{Male(z)}) \rightarrow \left ( x=y \vee x=z \vee y=z \right )$
for $0 \text{man}$ LHS will be false and whole implication will be true.
for $1 \text{man}$ LHS will be false and whole implication will be true.
for $2 \text{men}$ LHS will be false and whole implication will be true.
for >$2 \text{man}$ LHS will be true but RHS will be true iff all the men(>2) are just 2.
hence True $\rightarrow$ true=true and the whole implication will be true