A group of 3 is to be chosen from 7 boys and 3 girls. In how many ways can it be chosen if it must have at least one girl? (All 10 children are different.)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 85
"At least one" is found as all choices minus the choices with none.
All groups of 3 from 10: \(\displaystyle ^{10}C_3=\frac{10\times9\times8}{3\times2\times1}=120\)
Groups with no girl (3 boys from 7): \(\displaystyle ^7C_3=35\)
Groups with at least one girl \(=120-35=85\)
Why the other options are incorrect
Option (1) 120 — Incorrect. This is every group of 3, including the 35 groups of boys only.
Option (2) 35 — Incorrect. This is the number of groups with no girl.
Option (4) 108 — Incorrect. This is \(3\times\,^9C_2\): choosing "one girl first" counts a group with two or three girls more than once.