JEE Main 2026 Jan 23 Shift 2, Mathematics Q25: Combinations

Counting $n(S)$: A 4-digit number with strictly decreasing digits $a>b>c>d$ is determined by choosing any 4 distinct digits from $\{0,1,\dots ,9\}$ and arranging them in decreasing order. The leading digit is automatically non-zero.

$n(S)=\left(\genfrac{}{}{0ex}{}{10}{4}\right)=210$.

Counting $n(P)$: Need 5-digit ordered tuples (no leading zero, and no digit can be $0$ as that would zero the product) with digit product $=20$.

Factor $20=2^2\cdot 5$. Possible unordered digit multisets (each digit in $1$-$9$):

(i) $\{5,4,1,1,1\}$: arrangements $=\frac{5!}{3!}=20$.

(ii) $\{5,2,2,1,1\}$: arrangements $=\frac{5!}{2!2!}=30$.

(No other multisets work; for example $\{5,2,2,1,1\}$ exhausts the $2\cdot 2\cdot 5$ split.)

$n(P)=20+30=50$.

$\therefore n(S)+n(P)=210+50=260$.