JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Permutations and Combinations

For $n(S)$, any choice of $4$ distinct digits from $\{0,1,\dots ,9\}$ can be arranged in strictly decreasing order in exactly one way to give $a>b>c>d$, so $n(S)=\left(\genfrac{}{}{0ex}{}{10}{4}\right)=210$. For $n(P)$, the five digits multiply to $20=2^2\cdot 5$, forcing the digit multisets $\{5,4,1,1,1\}$, with $\frac{5!}{3!}=20$ arrangements, and $\{5,2,2,1,1\}$, with $\frac{5!}{2!2!}=30$ arrangements; thus $n(P)=20+30=50$. Therefore $n(S)+n(P)=210+50=260$.