JEE Main 2026 — 23 January, Evening Shift — Question 25

For set S: 4-digit numbers with digits $a>b>c>d$, digits from$0-9, a≠0.$ Choose any 4 distinct digits from 0-9: $\binom{10}{4}=210$ ways. Arrange them in descending order to get a unique number. So $n(S)=210$. For set P: 5-digit numbers with product of digits = 20, digits from 0-9, first digit ≠0. Possible multisets of digits: (5,4,1,1,1) and (5,2,2,1,1). For (5,4,1,1,1): number of distinct permutations = $\frac{5!}{3!}=20$. For (5,2,2,1,1): number of distinct permutations = $\frac{5!}{2!2!}=30$. Total $n(P)=20+30=50$. Hence $n(S)+n(P)=210+50=260$.