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

Introduction / Context:
This question evaluates your understanding of combinations and the complement method in probability style counting. Instead of directly counting all selections that contain at least one black ball, it is easier to subtract the cases with no black balls from the total number of ways to draw three balls.

Given Data / Assumptions:

• White balls = 2.
• Black balls = 3.
• Red balls = 4.
• Total balls = 2 + 3 + 4 = 9.
• We draw 3 balls at once.
• Condition: At least one ball among the three must be black.

Concept / Approach:
The total number of ways to choose 3 balls from 9 is 9C3. The only unwanted case is when none of the balls drawn is black. In that case, all three balls must come from the 2 white and 4 red balls, which is a total of 6 non black balls. So we calculate 6C3 and subtract it from 9C3 to get the desired count.

Step-by-Step Solution:
Step 1: Total ways to choose any 3 balls from 9 = 9C3.Compute 9C3 = 9 * 8 * 7 / 6 = 84.Step 2: Count ways with no black balls. Then we only use white and red balls: 2 white + 4 red = 6 balls.Number of ways to choose 3 non black balls = 6C3 = 6 * 5 * 4 / 6 = 20.Step 3: Favourable selections with at least one black ball = total selections - selections with no black ball.So required count = 84 - 20 = 64.

Verification / Alternative check:
You could also break into cases: exactly 1 black, exactly 2 black and exactly 3 black, and then add all counts.This alternative will also yield 64, but the complement method is more compact and avoids multiple case calculations.

Why Other Options Are Wrong:
48 and 45 are too small and can arise if one of the cases is missed.63 is just one less than the correct value and can result from an arithmetic slip such as mis computing 6C3.

Common Pitfalls:
Counting selections where at least one ball is black but forgetting to exclude the all non black scenario correctly.Miscomputing 9C3 or 6C3 due to factorial arithmetic errors.Mixing up order and combinations; here order does not matter, so combinations are correct.

Final Answer:
The number of ways to draw 3 balls with at least one black ball is 64.