The problem involves finding the number of valid passwords under the given restrictions. Here's the breakdown:

Information Extracted:

1. Characters: Each character is either:

   • A digit $[0-9]$ (10 options),
   • Or a lowercase letter $[a-z]$ (26 options).
   • Total: $10 + 26 = 36$ unique characters.

2. Password Length: $9$.

3. Conditions:

   • No character repeats.
   • Password must contain the character $a$.

Solution:

• Since $a$ must be included, choose $a$ to occupy 1 of the 9 positions. This can be done in $\binom{9}{1} = 9$ ways.

• Once $a$ is placed, there are $36 - 1 = 35$ remaining characters (excluding $a$), and $8$ positions left.

• Permutations for the other $8$ characters:

  • Select 8 characters from the remaining 35: $P(35, 8) = \frac{35!}{(35-8)!}$.

• Total number of valid passwords: $9 \times P(35, 8)$

Would you like the explicit calculation of $P(35, 8)$, or clarification of any part? Let me know!

Related Questions:

1. What if the password length changes to $n$? How does the formula generalize?
2. What happens if $a$ must appear multiple times in the password?
3. How does the solution change if repeats of characters are allowed?
4. Can this problem be solved using combinations instead of permutations?
5. What if certain characters (e.g., vowels) were not allowed in the password?

Tip:

Remember, permutations $P(n, k)$ emphasize the arrangement of characters, making them perfect for order-sensitive problems like passwords!