Let the 3-digit number be represented as \( abc \), where \( a, b, c \) are its digits.
The condition is:
\[ 2 < a \times b \times c < 7 \]
The possible values for the product \( a \times b \times c \) are:
\[ 3,\ 4,\ 5,\ 6 \]
The product is \( 1 \times 1 \times x = x \). Valid values for \( x \) are 3, 4, 5, 6. This yields 4 sets of digits:
Each set can form \( 3!/2! = 3 \) distinct numbers. Total numbers from this case: \( 4 \times 3 = 12 \).
The digits must be (2,2,1). The number of distinct arrangements is:
\[ \frac{3!}{2!} = 3 \]
The digits must be (1,2,3). Since all digits are distinct, the number of permutations is:
\[ 3! = 6 \]
The total number of valid 3-digit numbers is the sum of numbers from each case:
\[ 12 + 3 + 6 = \boxed{21} \]
21 valid numbers satisfy the condition \( 2 < a \times b \times c < 7 \).