## 1 Answer

Given that, the product of $3 \text{-digit}$ numbers is more than $2$ but less than $7.$

Let the $3\text{-digit}$ number be $xyz.$ Then, $ 2< x \times y \times z < 7 $

$ \Rightarrow x \times y \times z = 3\; \text{(or)}\; 4\;\text{ (or)}\; 5 \;\text{(or)}\; 6 $

$\textbf{Case 1:}$ When $x \times y \times z = 3$

- $\underline{x} \quad \underline{y} \quad \underline{z}$
- $1 \quad 1 \quad 3$
- $1 \quad 3 \quad 1$
- $3 \quad 1 \quad 1$

Three possibilities.

$\textbf{Case 2:}$ When $x \times y \times z = 4$

- $\underline{x} \quad \underline{y} \quad \underline{z}$
- $1 \quad 1 \quad 4$
- $1 \quad 4 \quad 1$
- $4 \quad 1 \quad 1$
- $1 \quad 2 \quad 2$
- $2 \quad 1 \quad 2$
- $2 \quad 2 \quad 1$

Six possibilities.

$\textbf{Case 3:}$ When $x \times y \times z = 5$

- $\underline{x} \quad \underline{y} \quad \underline{z}$
- $1 \quad 1 \quad 5$
- $1 \quad 5 \quad 1$
- $5 \quad 1 \quad 1$

Three possibilities.

$\textbf{Case 4:}$ When $x \times y \times z = 6$

- $\underline{x} \quad \underline{y} \quad \underline{z}$
- $1 \quad 2 \quad 3$
- $1 \quad 3 \quad 2$
- $2 \quad 3 \quad 1$
- $2 \quad 1 \quad 3$
- $3 \quad 1 \quad 2$
- $3 \quad 2 \quad 1$
- $1 \quad 1 \quad 6$
- $1 \quad 6 \quad 1$
- $6 \quad 1 \quad 1$

Nine possibilities.

Total numbers $ = 3 + 6 + 3 + 9 = 21.$

$\therefore$ There are $21$ numbers, whose product of their digits is more than $2$ but less than $7.$

Correct Answer$: 21$