**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.**

**Solution:**

We know that,

nCr

$=\frac{n !}{r !(n-r) !}$

(i) No girls

Total number of ways the team can have no girls = 4C0. 7C5 =21

(ii) at least one boy and one girl

Case(A)1 boy and 4 girls

=7C1. 4C4

=7

Case(B)2 boys and3 girls

=7C2. 4C3

=84

Case(C) 3boys and 2girls

=7C3. 4C2

=210

Case(D)4 boys and 1 girls

=7C4. 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 =4C3. 7C2+4C4.7C1

= 4 × 21 + 7

= 84 + 7

= 91