Understanding the Problem: Drawing Balls with a Condition

The question asks for the number of ways to select 3 balls from a box containing a mix of white, black, and red balls, with a specific condition: at least one of the selected balls must be black.

Here's the breakdown of the balls in the box:

• White balls: 2
• Black balls: 3
• Red balls: 4

The total number of balls in the box is $2 + 3 + 4 = 9$.

We need to find the number of ways to draw 3 balls such that our selection includes one or more black balls.

Strategy: Using the Complement Rule

The condition "at least one black ball" means we can have 1 black ball, 2 black balls, or 3 black balls in our selection of 3. Calculating each of these cases separately and summing them up is one way to solve this. However, a more efficient method is to use the complement rule:

Number of ways with at least one black ball = (Total number of ways to draw 3 balls) - (Number of ways to draw 3 balls with NO black balls)

Let's calculate each part:

Step 1: Calculate Total Number of Ways to Draw 3 Balls

We need to choose 3 balls from the total of 9 balls. This is a combination problem because the order in which we draw the balls does not matter. The number of ways to choose $k$ items from a set of $n$ items is given by the combination formula:

\(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

Here, $n=9$ (total balls) and $k=3$ (balls to draw). So, the total number of ways is:

\(\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!}\)

Let's expand the factorials:

\(\binom{9}{3} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(6 \times 5 \times 4 \times 3 \times 2 \times 1)}\)

Cancel out the $6!$ term:

\(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84\)

So, there are 84 total ways to draw 3 balls from the box.

Step 2: Calculate Number of Ways to Draw 3 Balls with NO Black Balls

To draw 3 balls with no black balls, we must select the 3 balls only from the non-black balls available. The non-black balls are the white and red balls.

• White balls: 2
• Red balls: 4

Total non-black balls: $2 + 4 = 6$.

We need to choose 3 balls from these 6 non-black balls. Again, this is a combination problem. Here, $n=6$ (non-black balls) and $k=3$ (balls to draw).

\(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}\)

Expand the factorials:

\(\binom{6}{3} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)}\)

Cancel out terms:

\(\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20\)

So, there are 20 ways to draw 3 balls that contain no black balls.

Step 3: Calculate Number of Ways with at Least One Black Ball

Using the complement rule:

Number of ways with at least one black ball = (Total ways to draw 3 balls) - (Ways to draw 3 balls with no black balls)

Number of ways with at least one black ball = $84 - 20 = 64$.

Therefore, there are 64 ways to draw 3 balls from the box with at least one black ball.

Conclusion

The number of ways of drawing 3 balls from the box with at least one black ball is 64.

Revision Table: Drawing Balls Combinations

Additional Information: Combinations vs. Permutations

It's important to distinguish between combinations and permutations in probability and counting problems.

• Combinations: Used when the order of selection does not matter. For example, choosing a committee of 3 people from a group of 10. Selecting John, Mary, and then Sam is the same as selecting Sam, Mary, and then John. The formula is \(\binom{n}{k}\).
• Permutations: Used when the order of selection *does* matter. For example, arranging 3 books on a shelf from a selection of 10. Putting book A, then B, then C is different from putting book C, then B, then A. The formula for permutations is \(P(n, k) = \frac{n!}{(n-k)!}\).

In this ball-drawing problem, since the question asks for the "number of ways of drawing 3 balls" without specifying any order or arrangement, it is a combination problem.

The "at least one" condition is a common trick in counting problems, and the complement rule (Total - None) is almost always the simplest way to solve it, especially when the "none" case is easy to calculate.