Question

A group consists of $4$ girls and $7$ boys. In how many on a team of $5$ members be selected, if the team has
(i) No girl.
(ii) at least one boy and one girl.
(iii) at least three girls.

Answer 1

No of girls $=4$

No of boys $=7$

Total member to be selected $=5$

(i)

No girl

No of girls to be selected $=0$

No of boys to be selected $=5$

No of ways in which girl can be selected $={}^4{C}_0$

No of ways in which boys can be selected $={}^7{C}_5$

No of ways team member can be selected with no girls $={}^4{C}_0\times {}^7{C}_5$

$=\frac{7!}{5!2!}=21$

(ii)

Atleast one boys and one girl

No of girls to be selected $=1,2,3,4$

No of boys to be selected $=1,2,3,4$

No of ways if there are $1$ girl and $4$ boys $={}^4{C}_1{}^7{C}_4$

No of ways if there are $2$ girls and $3$ boys $={}^4{C}_2{}^7{C}_3$

No of ways if there are $3$ girls and $2$ boys $={}^4{C}_3{}^7{C}_2$

No of ways of there are $4$ girls and $1$ boys $={}^4{C}_4{}^7{C}_1$

Total no of ways $={}^4{C}_1{}^7{C}_4+{}^4{C}_2{}^7{C}_3+{}^4{C}_3{}^7{C}_2+{}^4{C}_4{}^7{C}_1$

$=4\times 35+6\times 35+4\times 21+7$

$=140+210+84+7$

$=441$

(iii)

Atleast $3$ girls

No of girls to be selected $=3$ or $4$

No of boys to be selected $=2$ or $1$

Total is ways $={}^4{C}_3{}^7{C}_2+{}^4{C}_4{}^7{C}_1$

$=4\times 21+7=84+7=91$