(i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ⁷C₅ ways. Therefore, the required number of ways 

= ⁷C₅ (7!/5!2!) = (6 x 7)/2 = 21

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

(a) 1 boy and 4 girls 

(b) 2 boys and 3 girls 

(c) 3 boys and 2 girls 

(d) 4 boys and 1 girl. 

1 boys and 4 girls can be selected in ⁷C₁ × ⁴C₄ ways. 

2 boys and 3 girls can be selected in ⁷C₂ × ⁴C₃ ways. 

3 boys and 2 girls can be selected in ⁷C₃ × ⁴C₂ ways. 

4 boys and 1 girls can be selected in ⁷C₄ × ⁴C₁ ways. 

Therefore, the required number of ways = ⁷C₁ × ⁴C₄ + ⁷C₂ × ⁴C₃ + ⁷C₃ × ⁴C₂ + ⁷C₄ × ⁴C₁ 

= 7 + 84 + 210 + 140 = 441 

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

(a) 3 girls and 2 boys, or (b) 4 girls and 1 boy. 

Note that the team cannot have all 5 girls, because, the group has only 4 girls. 

3 girls and 2 boys can be selected in ⁴C₃ × ⁷C₂ ways. 

4 girls and 1 boys can be selected in ⁴C₄ × ⁷C₂ ways. 

Therefore, the required number of ways 

= ⁴C₃ × ⁷C₂ + ⁴C₄ × ₇C₁ = 84 + 7 =91.