JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Probability

There are $9\times 10^4$ five-digit numbers, since the leading digit cannot be zero. Count those built from exactly two distinct digits. Choosing two non-zero digits, in $\left(\genfrac{}{}{0ex}{}{9}{2}\right)$ ways, each fills the five places in $2^5-2=30$ ways once the two single-digit patterns are excluded. Choosing one non-zero digit together with zero, in $\left(\genfrac{}{}{0ex}{}{9}{1}\right)$ ways, the zero may not lead, leaving $2^4-1=15$ valid patterns. The favourable count is $36\times 30+9\times 15=1080+135=1215$, so the probability is $\frac{1215}{9\times 10^4}=\frac{135}{10^4}$.