There are 4 white, 3 black and 3 red 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 white, 3 black, and 3 red balls with the condition that at least one black ball must be included, we can break down the solution into clear steps.

### Step-by-Step Solution

1. **Identify the Total Balls**:
We have:
- 4 white balls
- 3 black balls
- 3 red balls
Total = 4 + 3 + 3 = 10 balls.

2. **Define the Cases**:
Since we need to select at least one black ball, we can break this down into three cases based on the number of black balls selected:
- Case 1: Selecting 1 black ball
- Case 2: Selecting 2 black balls
- Case 3: Selecting 3 black balls

3. **Case 1: Selecting 1 Black Ball**:
- Choose 1 black ball from 3: \( \binom{3}{1} \)
- Choose 2 balls from the remaining 7 (4 white + 3 red): \( \binom{7}{2} \)
- Total ways for this case:
\[ \binom{3}{1} \cdot \binom{7}{2} = 3 \cdot \frac{7 \times 6}{2 \times 1} = 3 \cdot 21 = 63 \]

4. **Case 2: Selecting 2 Black Balls**:
- Choose 2 black balls from 3: \( \binom{3}{2} \)
- Choose 1 ball from the remaining 7: \( \binom{7}{1} \)
- Total ways for this case:
\[ \binom{3}{2} \cdot \binom{7}{1} = 3 \cdot 7 = 21 \]

5. **Case 3: Selecting 3 Black Balls**:
- Choose all 3 black balls: \( \binom{3}{3} \)
- Choose 0 balls from the remaining 7: \( \binom{7}{0} \)
- Total ways for this case:
\[ \binom{3}{3} \cdot \binom{7}{0} = 1 \cdot 1 = 1 \]

6. **Total Ways**:
Now, we sum the total ways from all cases:
\[ \text{Total} = 63 + 21 + 1 = 85 \]

Thus, the number of ways to select three balls with at least one black ball is **85**.