JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Permutations and Combinations

We need to find 5-digit numbers whose digit product is 36. First, factorize $36=2^2\cdot 3^2$. The possible digit sets (using digits 1-9, with 1's as fillers) are: (1) $\{9,4,1,1,1\}$: arrangements $=\frac{5!}{3!}=20$. (2) $\{9,2,2,1,1\}$: arrangements $=\frac{5!}{2!2!}=30$. (3) $\{6,6,1,1,1\}$: arrangements $=\frac{5!}{2!3!}=10$. (4) $\{6,3,2,1,1\}$: arrangements $=\frac{5!}{2!}=60$. (5) $\{4,3,3,1,1\}$: arrangements $=\frac{5!}{2!2!}=30$. (6) $\{3,3,2,2,1\}$: arrangements $=\frac{5!}{2!2!}=30$. Total $=20+30+10+60+30+30=180$.