We need to construct 4-digit numbers using distinct digits from 0 to 9, ensuring that each number includes the digits 7 and 3.
We will analyze two scenarios:
The number structure is: 7 _ _ _
The digit 3 can be placed in any of the remaining 3 positions, which offers 3 distinct possibilities.
After placing 7 and 3, we have 8 digits left (excluding 7 and 3). The remaining two positions must be filled using these 8 digits. The number of ways to do this is:
\( ^8P_2 = 8 \times 7 = 56 \) ways
Therefore, the total count of numbers for this scenario is:
\( 3 \times 56 = 168 \)
The thousands place cannot be occupied by 0, 7, or 3. Thus, any of the remaining 7 digits can be placed in this position.
This can be done in 7 ways.
Next, the digits 7 and 3 must be placed in 2 of the 3 remaining positions. This is equivalent to selecting 2 positions out of 3 and arranging 7 and 3 in them:
\( ^3P_2 = 6 \)
Alternatively, we can select 2 positions for 7 and 3 in 3 ways, and then arrange 7 and 3 in those positions in 2! ways, resulting in \(3 \times 2 = 6\) ways to place 7 and 3 in the remaining 3 positions. A more direct way to think about placing 7 and 3 in the remaining 3 positions is to choose 2 positions, which can be done in \( \binom{3}{2} \) ways, and then arrange 7 and 3 in \( 2! \) ways, so \( \binom{3}{2} \times 2! = 3 \times 2 = 6 \) ways. Or simply, the first position can be filled in 3 ways and the second in 2 ways, thus \(3 \times 2 = 6\) ways.
The remaining 1 blank (after placing 7 and 3) can be filled from the remaining 7 digits (excluding the 3 digits already used), which can be done in:
\( 7 \) ways
Hence, the total number of distinct numbers in this scenario is:
\( 7 \times 6 \times 7 = 294 \)
The sum of numbers from both scenarios gives the total count:
\( 168 + 294 = \boxed{462} \)