Respuesta :
The selection of r objects out of n, can be done in C(n, r) many ways,
where [tex]\displaystyle{ C(n, r) = \frac{n!}{r!(n-r)!} [/tex], r! being [tex]1\cdot2\cdot...\cdot r[/tex].
Thus, 2 women out of 15 can be selected in a total of C(15, 2) many ways, and 2 men out of 12, can be selected in C(12, 2) many ways.
Any possible pair of women can be combined with any pair of men, thus there are a total of [tex]C(15, 2)\cdot C(12, 2)[/tex] many ways of forming the committee.
[tex]C(15, 2)\cdot C(12, 2)= \displaystyle{ \frac{15!}{2!13!}\cdot \frac{12!}{2!10!}= \displaystyle{ \frac{15\cdot14\cdot13!}{2!13!}\cdot \frac{12\cdot11\cdot10!}{2!10!}[/tex]
[tex]\displaystyle{ = \frac{15\cdot14}{2}\cdot \frac{12\cdot11}{2}=15\cdot 7\cdot6\cdot11= 6,930[/tex]
Answer: 6,930
where [tex]\displaystyle{ C(n, r) = \frac{n!}{r!(n-r)!} [/tex], r! being [tex]1\cdot2\cdot...\cdot r[/tex].
Thus, 2 women out of 15 can be selected in a total of C(15, 2) many ways, and 2 men out of 12, can be selected in C(12, 2) many ways.
Any possible pair of women can be combined with any pair of men, thus there are a total of [tex]C(15, 2)\cdot C(12, 2)[/tex] many ways of forming the committee.
[tex]C(15, 2)\cdot C(12, 2)= \displaystyle{ \frac{15!}{2!13!}\cdot \frac{12!}{2!10!}= \displaystyle{ \frac{15\cdot14\cdot13!}{2!13!}\cdot \frac{12\cdot11\cdot10!}{2!10!}[/tex]
[tex]\displaystyle{ = \frac{15\cdot14}{2}\cdot \frac{12\cdot11}{2}=15\cdot 7\cdot6\cdot11= 6,930[/tex]
Answer: 6,930
