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 girls

(ii) at least one boy and one girl

(iii) at least three girls.

Answer 1

We know that,

(i) No girls

Total number of ways the team can have no girls = 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 the team can have at least one boy and one girl,

= Case(A) + Case(B) + Case(C) + Case(D)

=7 + 84 + 210 + 140

=441

(iii) At least three girls

Total number of ways the team can have at least three girls =⁴C_3. ⁷C₂+⁴C_4.⁷C₁

= 4 × 21 + 7

= 84 + 7

= 91