There are `n-1` red balls, n green balls and `n+1` blue balls in a bag. The number of ways of choosing two balls from the bag that have different colours is 299. What is the value of n?

Answer

To solve the problem, we need to find the value of \( n \) given that the number of ways to choose two balls of different colors from a bag containing \( n-1 \) red balls, \( n \) green balls, and \( n+1 \) blue balls is 299.

### Step-by-Step Solution:

1. **Identify the number of balls of each color**:
- Red balls: \( n - 1 \)
- Green balls: \( n \)
- Blue balls: \( n + 1 \)

2. **Determine the combinations of choosing two balls of different colors**:
- **Case 1**: Choosing 1 red and 1 green:
\[ \text{Ways} = (n - 1) \cdot n \]
- **Case 2**: Choosing 1 green and 1 blue:
\[ \text{Ways} = n \cdot (n + 1) \]
- **Case 3**: Choosing 1 red and 1 blue:
\[ \text{Ways} = (n - 1) \cdot (n + 1) \]

3. **Combine the cases**:
The total number of ways to choose two balls of different colors is the sum of the ways from all three cases:
\[ \text{Total Ways} = (n - 1)n + n(n + 1) + (n - 1)(n + 1) \]

4. **Expand and simplify the expression**:
- For Case 1:
\[ (n - 1)n = n^2 - n \]
- For Case 2:
\[ n(n + 1) = n^2 + n \]
- For Case 3:
\[ (n - 1)(n + 1) = n^2 - 1 \]
- Now, combine all these:
\[ \text{Total Ways} = (n^2 - n) + (n^2 + n) + (n^2 - 1) \]
- Combine like terms:
\[ \text{Total Ways} = 3n^2 - 1 \]

5. **Set the total ways equal to 299**:
\[ 3n^2 - 1 = 299 \]

6. **Solve for \( n \)**:
- Add 1 to both sides:
\[ 3n^2 = 300 \]
- Divide by 3:
\[ n^2 = 100 \]
- Take the square root:
\[ n = 10 \quad (\text{since } n \text{ must be positive}) \]

### Final Answer:
The value of \( n \) is \( 10 \).