Question

From the first 100 natural numbers, two numbers first \( a \) and then \( b \) are selected randomly without replacement. If the probability that \( a - b \ge 10 \) is \( m/n \), \( \text{gcd}(m, n) = 1 \), then \( m + n \) is equal to :

Hint:

For "without replacement" problems where order matters (first \(a\), then \(b\)), the total sample space is $n(n-1)$. If order didn't matter, it would be $^nC_2$.

Answer 1

Correct Answer: 311

Step 1: Understanding the Concept:
We need to count the number of pairs \((a, b)\) such that \( 1 \le a, b \le 100 \), \( a \neq b \), and \( a - b \ge 10 \).
Step 2: Key Formula or Approach:
Total outcomes: \( 100 \times 99 = 9900 \). Favorable outcomes: \( a \ge b + 10 \).
Step 3: Detailed Explanation:
- If \( b = 1 \), \( a \in \{11, 12, \dots, 100\} \) (90 values). - If \( b = 2 \), \( a \in \{12, 13, \dots, 100\} \) (89 values). - ... - If \( b = 90 \), \( a \in \{100\} \) (1 value). This is an AP: \( 1 + 2 + \dots + 90 \). Sum \( = \frac{90 \times 91}{2} = 45 \times 91 = 4095 \). Probability \( P = \frac{4095}{9900} \). Divide by 45: \( \frac{91}{220} \). Here \( m = 91, n = 220 \). \( \text{gcd}(91, 220) = 1 \). \( m + n = 91 + 220 = 311 \).
Step 4: Final Answer:
The value is 311.