Resposta :
Circunstância em que há 5 mulheres (portanto 2 homens):
[tex]C_{5,5} \cdot C_{8,2} \\\\= \cfrac{5!}{5!} \cdot \cfrac{8!}{6!2!}\\\\ = 1 \cdot \cfrac{8 \cdot 7 \cdot 6!}{6! \cdot 2} \\\\= 4 \cdot 7\\= 28[/tex]
Com 4 mulheres (3 homens):
[tex]C_{5,4} \cdot C_{8,3} \\\\= \cfrac{5!}{4! \cdot 1!} \cdot \cfrac{8!}{5!3!}\\\\ = 5 \cdot \cfrac{8 \cdot 7 \cdot 6 \cdot 5!}{5! \cdot 6} \\\\= 8 \cdot 7\\= 56[/tex]
Com 3 mulheres (4 homens):
[tex]C_{5,3} \cdot C_{8,4} \\\\= \cfrac{5!}{3! \cdot 2!} \cdot \cfrac{8!}{4!4!}\\\\ = \cfrac{5 \cdot4\cdot 3!}{3! \cdot 2} \cdot \cfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{4! \cdot 8 \cdot 3} \\\\= 5 \cdot 2 \cdot 7 \cdot 3 \cdot 5\\= 1050[/tex]
Somando as possibilidades:
[tex]28 + 56 + 1050 = 1134[/tex]