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From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $\mathrm{a}-\mathrm{b} \geq 10$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$ .[JEE MAIN 23 Jan 2026 S I]
Solution
$\mathrm{a}, \mathrm{b} \in\{1,2,3, \ldots, 99,100\}$
Total cases $=100 \times 99$
Favourable cases $a-b \geq 10$
No of Pairs of (a,b)\[ \begin{array}{ll} b=1, a \geq 11 & a \rightarrow 11,12,13, \ldots, 99,100 \\ b=2, a \geq 12 & a \rightarrow 12,13,14, \ldots, 99,100 \end{array} \]$b=3, a \geq 13$ $a \rightarrow 13,14,15, \ldots, 99,100$ 88 $\vdots$ $b=88, a \geq 98$ $a \rightarrow 98,99,100$ 3 $b=89, a \geq 99$ $a \rightarrow 99,100$ 2 $b=90, a \geq 100$ $a \rightarrow 100$ 1 No. of favourable ordered pair $=1+2+3+\ldots .+89+90$
\[ =\frac{90(90+1)}{2}=45 \times 91 \]Required probability $=\frac{45 \times 91}{100 \times 99}=\frac{91}{220}=\frac{\mathrm{m}}{\mathrm{n}}$
\[ \mathrm{m}+\mathrm{n}=91+220=311 \text { Answer }(311) \]
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