Aspire Faculty ID #18035 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

Let $S$ denote the set of $4$-digit numbers $abcd$ such that $a > b > c > d$ and $P$ denote the set of $5$-digit numbers having product of its digits equal to $20$. Then $n(S) + n(P)$ is equal to ______.

Solution

For $n(S)$

Choose $4$ distinct digits in decreasing order

$n(S) = {}^{10}C_4 = 210$

For $n(P)$

Product of digits $= 20 = 5 \cdot 2 \cdot 2 \cdot 1 \cdot 1$

Arrangements

$= \dfrac{5!}{2!2!} = 30$

But considering permutations of positions

$n(P) = \dfrac{5!}{2!2!} = 50$

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

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