Two different numbers are selected at random from the set S={1, 2, 3, ……10}, then the probability that sum of selected numbers is divisible by 2 a/b, where a and b are co-prime then b is equal to ___.

Answer

To solve the problem, we need to find the probability that the sum of two different numbers selected from the set \( S = \{1, 2, 3, \ldots, 10\} \) is divisible by 2.

### Step-by-Step Solution:

1. **Identify the total number of ways to select two different numbers from the set \( S \)**:
The total number of ways to choose 2 numbers from 10 is given by the combination formula:
\[ \text{Total ways} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]

**Hint**: Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to find the number of ways to choose \( r \) items from \( n \) items.

2. **Determine the cases where the sum of the two numbers is divisible by 2**:
A sum is divisible by 2 if both numbers are either even or both are odd.

- **Case 1**: Both numbers are even.
The even numbers in the set \( S \) are \( \{2, 4, 6, 8, 10\} \). There are 5 even numbers.
The number of ways to choose 2 even numbers:
\[ \text{Ways (even)} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]

- **Case 2**: Both numbers are odd.
The odd numbers in the set \( S \) are \( \{1, 3, 5, 7, 9\} \). There are also 5 odd numbers.
The number of ways to choose 2 odd numbers:
\[ \text{Ways (odd)} = \binom{5}{2} = 10 \]

3. **Calculate the total favorable outcomes**:
The total number of favorable outcomes (both numbers even or both numbers odd):
\[ \text{Total favorable outcomes} = \text{Ways (even)} + \text{Ways (odd)} = 10 + 10 = 20 \]

**Hint**: Add the favorable outcomes from both cases to get the total number of favorable outcomes.

4. **Calculate the probability**:
The probability that the sum of the selected numbers is divisible by 2 is given by:
\[ P(\text{sum divisible by 2}) = \frac{\text{Total favorable outcomes}}{\text{Total ways}} = \frac{20}{45} \]

5. **Simplify the probability**:
Simplifying \( \frac{20}{45} \):
\[ \frac{20}{45} = \frac{4}{9} \]

Here, \( a = 4 \) and \( b = 9 \). Since 4 and 9 are co-prime (they have no common factors other than 1), we can conclude that \( b \) is equal to 9.

### Final Answer:
The value of \( b \) is \( \boxed{9} \).