A box contains 2 white, 3 black, and 4 red balls. In how many ways can 3 balls be drawn if at least one black ball must be included?

Problem restatementCount unordered draws of 3 balls from 9 distinct balls, with the condition that at least one is black.

Given data

• Total balls = 2 (white) + 3 (black) + 4 (red) = 9.
• Draws are combinations (order does not matter).
• Constraint: include at least one black ball.

Concept/ApproachUse complement counting: total 3-ball combinations minus those with zero black balls.

Step-by-step calculation Total ways = C(9, 3) = 84 Zero-black ways = choose from 2W + 4R = 6 balls ⇒ C(6, 3) = 20 Required ways = 84 − 20 = 64

Verification/AlternativeDirect sum over cases with 1B, 2B, 3B yields the same count: C(3,1)C(6,2) + C(3,2)C(6,1) + C(3,3)C(6,0) = 3×15 + 3×6 + 1×1 = 45 + 18 + 1 = 64.

Common pitfalls

• Treating balls as indistinguishable by color; here each physical ball is distinct, so use standard combinations.

Final Answer64