Condição I: mulheres
[tex]C_{m,2}=21\\\\\frac{m!}{(m-2!)2!}=21\\\\\frac{m\cdot(m-1)\cdot(m-2)!}{(m-2!)2\cdot1}=21\\\\m(m-1)=42\\m^2-m-42=0\\(m-7)(m+6)=0\\\boxed{m=7}[/tex]
Condição II:
- duas mulheres,
[tex]C_{m,2}\cdot n=\\\\\frac{7!}{(7-2!)2!}\cdot n=\\\\\frac{7\cdot6\cdot5!}{5!2\cdot1}\cdot n=\boxed{21n}[/tex]
- três mulheres,
[tex]C_{m,3}=\\\\\frac{7!}{(7-3!)3!}=\\\\\frac{7\cdot6\cdot5\cdot4!}{4!3\cdot2\cdot1}=\boxed{35}[/tex]
Segue,
[tex]21n+35=140\\21n=105\\\boxed{n=5}[/tex]
Logo,
[tex]m+n=7+5\\\boxed{\boxed{m+n=12\;\text{pessoas}}}[/tex]