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 girls

(ii) at least one boy and one girl

(iii) at least three girls.

Formula:- (i)ⁿC_r

(i) No girls

Total number of ways

=⁴C_0. ⁷C₅

=21

(ii) at least one boy and one girl

CASE(A)1 boy and 4 girls

=⁷C_1. ⁴C₄

=7

CASE(B)2 boys and3 girls

=⁷C_2. ⁴C₃

=84

CASE(C) 3boys and 2girls

=⁷C_3. ⁴C₂

=210

CASE (D)4 boys and 1 girls

=⁷C_4. ⁴C₁

=140

Total number of ways= CASE(A)+CASE(B)+CASE(C)+CASE(D)

=7+84+210+140

=441

(iii) At least three girls

=⁴C_3. ⁷C₂+⁴C_4.⁷C₁

= 4 × 21 + 7

= 84 + 7

= 91