Reply To: Question 46 — 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 $=\frac{45 \times 91}{100 \times 99}=\frac{91}{220}=\frac{\mathrm{m}}{\mathrm{n}}$
\[
\mathrm{m}+\mathrm{n}=91+220=311 \text { Answer }(311)
\]