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 Set and Total Outcomes**:
The set \( A \) contains 10 elements. We need to select 3 different numbers from this set. The total number of ways to choose 3 numbers from 10 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Total outcomes} = \binom{10}{3} \]

2. **Calculate \( \binom{10}{3} \)**:
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 \]

3. **Identify Favorable Outcomes**:
We need to find the sets of three numbers \( (a, b, c) \) such that \( ab = c \). We can systematically check combinations:
- \( (2, 3, 6) \) since \( 2 \times 3 = 6 \)
- \( (2, 4, 8) \) since \( 2 \times 4 = 8 \)
- \( (2, 5, 10) \) since \( 2 \times 5 = 10 \)

Thus, the favorable outcomes are:
- \( (2, 3, 6) \)
- \( (2, 4, 8) \)
- \( (2, 5, 10) \)

Therefore, the number of favorable outcomes is 3.

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

5. **Identify \( p \) and \( q \)**:
From the probability \( \frac{1}{40} \), we have \( p = 1 \) and \( q = 40 \).

6. **Calculate \( p + q \)**:
\[ p + q = 1 + 40 = 41 \]

### Final Answer:
Thus, the value of \( p + q \) is \( \boxed{41} \).