There are 4 red, 3 black and 5 white balls in a bag. Find the number of ways of selecting three balls, if at least one black ball is there.

Answer

To solve the problem of selecting three balls from a bag containing 4 red, 3 black, and 5 white balls with the condition that at least one black ball must be included, we can break down the solution into several steps.

### Step-by-Step Solution:

1. **Understand the Total Balls**:
- We have a total of 4 red, 3 black, and 5 white balls.
- Total number of balls = 4 + 3 + 5 = 12 balls.

2. **Identify Cases for Selection**:
- Since we need to select at least one black ball, we can break this down into three cases:
- Case 1: Selecting 1 black ball, and 2 from the remaining (red + white).
- Case 2: Selecting 2 black balls, and 1 from the remaining (red + white).
- Case 3: Selecting 3 black balls.

3. **Calculate Each Case**:
- **Case 1**: Selecting 1 black ball and 2 from the remaining 9 balls (4 red + 5 white).
- Number of ways to choose 1 black ball from 3: \( \binom{3}{1} = 3 \)
- Number of ways to choose 2 balls from 9: \( \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \)
- Total for Case 1: \( 3 \times 36 = 108 \)

- **Case 2**: Selecting 2 black balls and 1 from the remaining 9 balls.
- Number of ways to choose 2 black balls from 3: \( \binom{3}{2} = 3 \)
- Number of ways to choose 1 ball from 9: \( \binom{9}{1} = 9 \)
- Total for Case 2: \( 3 \times 9 = 27 \)

- **Case 3**: Selecting 3 black balls.
- Number of ways to choose 3 black balls from 3: \( \binom{3}{3} = 1 \)
- Number of ways to choose 0 from the remaining 9: \( \binom{9}{0} = 1 \)
- Total for Case 3: \( 1 \times 1 = 1 \)

4. **Combine All Cases**:
- Total number of ways to select the balls with at least one black ball:
\[ \text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} = 108 + 27 + 1 = 136 \]

### Final Answer:
The total number of ways to select three balls, ensuring at least one black ball is included, is **136**.

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