The problem asks us to calculate the total number of valid passwords that meet the following conditions:

1. Password length is 12 characters.
2. Characters can only be digits (0–9) or lowercase letters (a–z).
3. No character repeats.
4. The password must contain the specific characters a, 4, 7, and 3.

Step-by-Step Solution:

Step 1: Total pool of characters

The characters consist of:

• 10 digits (0–9)
• 26 lowercase letters (a–z)

This gives a total of $10 + 26 = 36$ unique characters.

Step 2: Fix the required characters

The required characters (a, 4, 7, and 3) must appear in the password. That accounts for 4 fixed characters.

Step 3: Choose the remaining characters

Since the total password length is 12 and 4 characters are fixed, there are $12 - 4 = 8$ characters left to be chosen from the remaining pool of characters. After removing the required 4 characters, the pool contains $36 - 4 = 32$ characters.

We need to select 8 characters from this pool of 32. The number of ways to choose and arrange 8 characters is given by the permutation formula, $P(n, k)$, which represents the number of ways to arrange $k$ items from $n$ options: $P(32, 8) = \frac{32!}{(32 - 8)!}$

Step 4: Arrange all 12 characters

The 12 characters (4 fixed and 8 selected) can be arranged in any order. The number of permutations for arranging 12 items is: $P(12, 12) = 12!$

Step 5: Combine results

The total number of valid passwords is: $P(32, 8) \times P(12, 12)$

We substitute the values into the formula for permutations: $P(n, k) = \frac{n!}{(n - k)!}$

• $P(32, 8) = \frac{32!}{24!}$
• $P(12, 12) = 12!$

This gives the final calculation for the total number of valid passwords.

Would you like me to compute the exact numerical value or clarify any part of this solution?