In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Solution:
We may have,
(1 boy and 3 girls) = (⁶C₁ x ⁴C₃)
= 6 × 4 ways
= 24 ways

(2 boys and 2 girls) = (⁶C₂ x ⁴C₂)
= 15 × 6 ways
= 90 ways

(3 boys and 1 girl) = (⁶C₃ x ⁴C₁)
= 20 × 4 ways
= 80 ways

(4 boys) = ⁶C₂ = 15 ways

∴ Required number of ways
= (24 + 90 + 80 + 15) ways
= 209 ways