From the first 100 natural numbers, two numbe | vedclass

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(B) The total number of ways to select two distinct numbers $a$ and $b$ from $100$ natural numbers is $100 	imes 99 = 9900$.We want to find the numbe

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$311$

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(B) The total number of ways to select two distinct numbers $a$ and $b$ from $100$ natural numbers is $100 	imes 99 = 9900$.We want to find the number of pairs $(a, b)$ such that $a-b \ge 10$, which implies $a \ge b+10$.If $b=1$, $a$ can be any value from $11$ to $100$ ($90$ values).If $b=2$, $a$ can be any value from $12$ to $100$ ($89$ values).Continuing this, if $b=90$, $a$ can only be $100$ ($1$ value).Total favorable cases $= 90 + 89 + \dots + 1 = \frac{90 	imes 91}{2} = 4095$.The probability is $\frac{4095}{9900}$.Dividing both numerator and denominator by their greatest common divisor, $45$, we get $\frac{4095 \div 45}{9900 \div 45} = \frac{91}{220}$.Here, $m=91$ and $n=220$, so $\gcd(91, 220) = 1$.Therefore, $m+n = 91 + 220 = 311$.

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

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