Question 13

Explanation

We will determine the number of possible passwords using the multiplication rule of combinatorics.

Password structure: Digit - Letter - Digit - Letter (starts with a digit and alternates).

Counting Choices for Position 1 (First Digit)

The first digit is an odd number ($1, 3, 5, 7, 9$).

Number of choices = $5$.

Counting Choices for Position 2 (First Letter)

Vowels: $A, E, I, O, U$ (total $5$ vowels).

The first letter is not the letter $A$, so the remaining choices are: $E, I, O, U$.

Number of choices = $4$.

Counting Choices for Position 3 (Second Digit)

The second digit is chosen from $2, 4, 6, 8$ (total $4$ choices).

Since digits cannot repeat, we must ensure the second digit is different from the first digit. However, the first digit is odd and the second digit is even, so they automatically do not repeat.

Number of choices = $4$.

Counting Choices for Position 4 (Second Letter)

The second letter is the letter $O$ (fixed).

Number of choices = $1$.

Calculating Total Passwords

Using the multiplication rule:

However, we need to check if letters can repeat. From the problem, letters cannot repeat. Let's check the case where the first letter is $O$ (same as the second letter).

If the first letter = $O$, then there is letter repetition. Valid choices for the first letter are $E, I, U$ (cannot be $A$ and cannot be $O$ because $O$ is already used in the second position).

So:

• First digit = $5$ choices.
• First letter = $3$ choices ($E, I, U$).
• Second digit = $4$ choices.
• Second letter = $1$ choice ($O$).

So, the number of passwords that can be formed is $60$.