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