Three different numbers are selected at random from the set `A={1,2,3,……..,10}`. Then the probability that the product of two numbers equal to the third number is `(p)/(q)`, where p and q are relatively prime positive integers then the value of `(p+q)` is :

Options

A. 39
B. 40
C. 41
D. 42

Answer

To solve the problem, we need to find the probability that the product of two numbers equals the third number when selecting three different numbers from the set \( A = \{1, 2, 3, \ldots, 10\} \).

### Step-by-Step Solution:

1. **Identify the Condition**: We need to find three different numbers \( x, y, z \) such that \( x \cdot y = z \).

2. **Consider the Possibilities**:
- The smallest number we can use is 2 (since using 1 would lead to repetition).
- Let's explore pairs of numbers starting from 2.

3. **Case Analysis**:
- **Case 1**: Let \( x = 2 \).
- \( y = 3 \) gives \( z = 2 \cdot 3 = 6 \) → Valid triplet: \( (2, 3, 6) \).
- \( y = 4 \) gives \( z = 2 \cdot 4 = 8 \) → Valid triplet: \( (2, 4, 8) \).
- \( y = 5 \) gives \( z = 2 \cdot 5 = 10 \) → Valid triplet: \( (2, 5, 10) \).
- **Case 2**: Let \( x = 3 \).
- \( y = 2 \) gives \( z = 3 \cdot 2 = 6 \) → Already counted.
- \( y = 4 \) gives \( z = 3 \cdot 4 = 12 \) → Out of range.
- **Case 3**: Let \( x = 4 \).
- \( y = 2 \) gives \( z = 4 \cdot 2 = 8 \) → Already counted.
- \( y = 3 \) gives \( z = 4 \cdot 3 = 12 \) → Out of range.
- **Case 4**: Let \( x = 5 \).
- \( y = 2 \) gives \( z = 5 \cdot 2 = 10 \) → Already counted.
- \( y = 3 \) gives \( z = 5 \cdot 3 = 15 \) → Out of range.
- **Case 5**: Let \( x = 6 \).
- All products exceed 10.
- **Case 6**: Let \( x = 7, 8, 9, 10 \).
- All products exceed 10.

4. **Count Favorable Cases**:
- The valid triplets we found are: \( (2, 3, 6), (2, 4, 8), (2, 5, 10) \).
- Total favorable cases = 3.

5. **Calculate Total Cases**:
- The total ways to choose 3 different numbers from 10 is given by \( \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \).

6. **Calculate Probability**:
- Probability \( P \) that the product of two numbers equals the third number is given by:
\[ P = \frac{\text{Number of favorable cases}}{\text{Total cases}} = \frac{3}{120} = \frac{1}{40}. \]

7. **Identify \( p \) and \( q \)**:
- Here, \( p = 1 \) and \( q = 40 \).
- Since \( p \) and \( q \) are relatively prime, we can now find \( p + q \).

8. **Final Calculation**:
- \( p + q = 1 + 40 = 41 \).

### Conclusion:
The value of \( p + q \) is \( \boxed{41} \).