A box contains 2 red balls, 3 black balls and 4 white balls, all different from one another. In how many ways can 3 balls be drawn from the box so that at least 1 black ball is drawn?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 64
For "at least one", it is quickest to take all selections minus the selections with none of that kind. The balls are treated as different, so each count is a combination.
Total balls \(=2+3+4=9\)
All ways to draw 3 balls: \(^9C_3=84\)
Ways with no black ball (3 from the 6 red and white): \(^6C_3=20\)
At least 1 black \(=84-20=64\)
Why the other options are incorrect
Option (2) 48 — Incorrect. The cases 1, 2 and 3 black give \(45+18+1=64\), not 48.
Option (3) 32 — Incorrect. This is only half of the 64 selections that contain a black ball.
Option (4) 96 — Incorrect. It is more than all 84 possible selections of 3 balls, so it cannot be right.