Question 36 — JEE Main 23 Jan 2026 S Ii

Question

Let S denote the set of 4-digit numbers abcd such that $\mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{d}$ and P denote the set of 5 -digit numbers having product of its digits equal to 20 . Then $\mathrm{n}(\mathrm{S})+\mathrm{n}(\mathrm{P})$ is equal to $\_\_\_\_$ [JEE MAIN 23 Jan 2026 S II]

Solution

0,1,2,3,4,5,6,7,8,9
$\mathrm{n}(\mathrm{S})={ }^{10} \mathrm{C}_{4} \times 1=\frac{10 \cdot 9 \cdot 8 \cdot 7}{24}=210$
xyzwt $=20$
Case I: 5, 4, 1, 1, 1
Such numbers formed $=\frac{5!}{3!}=\frac{120}{6}=20$
Case II: 5, 2, 2, 1, 1
Such numbers formed $=\frac{5!}{2!2!}=\frac{120}{4}=30$
$\mathrm{n}(\mathrm{P})=20+30=50$
$\mathrm{n}(\mathrm{s})+\mathrm{n}(\mathrm{p})=210+50=260$ answer(260)