Q50 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $a-b \geqslant 10$ is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to $\_\_\_\_$.
Correct answer: 311
Solution
Total ordered selections $= 100 \times 99 = 9900$.
Favorable ($a - b \geq 10$): for each $b$ from 1 to 90, $a$ ranges from $b+10$ to 100, giving $91 - b$ choices.
Count $= \sum_{b=1}^{90}(91-b) = \sum_{k=1}^{90}k = \frac{90 \times 91}{2} = 4095$.
Probability $= \frac{4095}{9900} = \frac{91}{220}$ (after dividing by 45).
$\gcd(91, 220) = 1$, so $m = 91, n = 220$.
$m + n = 311$.
Favorable ($a - b \geq 10$): for each $b$ from 1 to 90, $a$ ranges from $b+10$ to 100, giving $91 - b$ choices.
Count $= \sum_{b=1}^{90}(91-b) = \sum_{k=1}^{90}k = \frac{90 \times 91}{2} = 4095$.
Probability $= \frac{4095}{9900} = \frac{91}{220}$ (after dividing by 45).
$\gcd(91, 220) = 1$, so $m = 91, n = 220$.
$m + n = 311$.
Hence, the correct answer is 311.