How many 8-digit numbers are there such that exactly 7 digits are all different?

$\textbf{(A)}\  {{9}\choose{3}}^2 \cdot 6! \cdot 3 \qquad\textbf{(B)}\  {{8}\choose{3}}^2 \cdot 7! \qquad\textbf{(C)}\  {{8}\choose{3}}^2 \cdot 7! \cdot 3 \\ \qquad\textbf{(D)}\  {{7}\choose{3}}^2 \cdot 7! \qquad\textbf{(E)}\  {{9}\choose{4}}^2 \cdot 6! \cdot 8$
1. **Selecting the first digit:**
   We can select the first digit in \( \binom{9}{1} \) ways. This is because we have 9 possible digits (1 through 9) to choose from for the first digit.

2. **Selecting the remaining 7 digits:**
   For the remaining 7 digits, we need 6 different digits (since one digit will be repeated). We can choose these 6 different digits from the remaining 8 digits (0 through 9, excluding the first digit) in \( \binom{8}{6} \) ways.

3. **Choosing the repeated digit:**
   The repeated digit can either be the first digit or one of the 6 different digits chosen in step 2. If the repeated digit is the first digit, then all 7 remaining digits are different. If the repeated digit is one of the 6 different digits, we need to choose which one of the 6 will be repeated.

4. **Calculating the permutations:**
   - If the repeated digit is the first digit, then we have 7! permutations of the remaining 7 digits.
   - If the repeated digit is one of the 6 different digits, we choose the repeated digit in \( \binom{6}{1} \) ways and then arrange the 7 digits (with one digit repeated) in \( \frac{7!}{2!} \) ways (since one digit is repeated twice).

5. **Combining the cases:**
   - The total number of permutations if the repeated digit is the first digit: \( 7! \)
   - The total number of permutations if the repeated digit is one of the 6 different digits: \( \binom{6}{1} \cdot \frac{7!}{2!} \)

6. **Summing the permutations:**
   The total number of permutations is:
   \[
   7! + \binom{6}{1} \cdot \frac{7!}{2!}
   \]

7. **Final calculation:**
   - The number of ways to choose the first digit: \( \binom{9}{1} \)
   - The number of ways to choose the 6 different digits: \( \binom{8}{6} \)
   - The total number of permutations: \( 7! + \binom{6}{1} \cdot \frac{7!}{2!} \)

   Therefore, the total number of 8-digit numbers is:
   \[
   \binom{9}{1} \cdot \binom{8}{6} \cdot \left( 7! + \binom{6}{1} \cdot \frac{7!}{2!} \right)
   \]

8. **Simplifying the expression:**
   \[
   \binom{9}{1} = 9
   \]
   \[
   \binom{8}{6} = \binom{8}{2} = 28
   \]
   \[
   7! = 5040
   \]
   \[
   \binom{6}{1} = 6
   \]
   \[
   \frac{7!}{2!} = \frac{5040}{2} = 2520
   \]
   \[
   7! + \binom{6}{1} \cdot \frac{7!}{2!} = 5040 + 6 \cdot 2520 = 5040 + 15120 = 20160
   \]
   \[
   \binom{9}{1} \cdot \binom{8}{6} \cdot \left( 7! + \binom{6}{1} \cdot \frac{7!}{2!} \right) = 9 \cdot 28 \cdot 20160
   \]

9. **Final expression:**
   \[
   9 \cdot 28 \cdot 20160 = 9 \cdot 28 \cdot 20160 = 9 \cdot 564480 = 5080320
   \]

Thus, the correct answer is:
\[
\boxed{A}
\]