Product of digits of 3-digit numbers

We need to find the number of 3-digit numbers whose digit product $P$ satisfies $2 < P < 7$, i.e., $P \in \{3, 4, 5, 6\}$.

Let the 3-digit number be $abc$, where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$. Since $P > 2$, none of the digits can be 0.

• Case 1: Product = 3 Digits must be a permutation of $\{1, 1, 3\}$. Number of arrangements $= \frac{3!}{2!} = 3$.

• Case 2: Product = 4

  • Set $\{1, 1, 4\}$: $\frac{3!}{2!} = 3$ numbers.
  • Set $\{1, 2, 2\}$: $\frac{3!}{2!} = 3$ numbers. Total for Product 4 $= 3 + 3 = 6$.

• Case 3: Product = 5 Digits must be a permutation of $\{1, 1, 5\}$. Number of arrangements $= \frac{3!}{2!} = 3$.

• Case 4: Product = 6

  • Set $\{1, 1, 6\}$: $\frac{3!}{2!} = 3$ numbers.
  • Set $\{1, 2, 3\}$: $3! = 6$ numbers. Total for Product 6 $= 3 + 6 = 9$.

Summing all cases: $3 + 6 + 3 + 9 = 21$.