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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

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