Question

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ?

(i) at least one boy and one girl ?

(iii) at least 3 girls ?

Answer 1

(i) Since, the team does not include any girl therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ${}^7{C}_5$ ways.

${}^7{C}_5=\frac{7!}{5!2!}=\frac{7\times 6}{2}=21$

(ii) Since, at least one boy and one girl are to be there in every team. The team consist of

(a) 1 boy and 4 girls i.e. ${}^7{C}_1{\times }^4{C}_4$

(b) 2 boy and 3 girls i.e. ${}^7{C}_2{\times }^4{C}_3$

(c) 3 boy and 2 girls i.e. ${}^7{C}_3{\times }^4{C}_2$

(d) 4 boy and 1 girls i.e. ${}^7{C}_4{\times }^4{C}_1$

$\therefore$ The required number of ways

${=}^7{C}_1{\times }^4{C}_4{\times }^7{C}_2{\times }^4{C}_3{+}^7{C}_3{\times }^4{C}_2{+}^7{C}_4{\times }^4{C}_1$

$=7+84+210+140$

$=441$

(iii) Since, the team has to conisst of at least 3 girls, the team can consist of

(a) 3 girls and 2 boys ${=}^7{C}_2{\times }^4{C}_3$ ways

(b) 4 girls and 1 boys ${=}^4{C}_4{\times }^7{C}_1,$ ways

$\therefore$ The required number of ways

${=}^7{C}_2{\times }^4{C}_3{+}^4{C}_4{\times }^7{C}_1$

$=84+7=91$