To tackle this exercise systematically, let's break down each part. The password has specific requirements:

• It is composed of 6 characters.
• The structure is as follows: 2 vowels + 2 even digits + 2 letters.
• Repetitions are not allowed.

Assumptions

1. Vowels: We assume the French vowels "A, E, I, O, U," providing 5 possible choices.
2. Even digits: We assume the even digits to be "0, 2, 4, 6, 8," providing 5 possible choices.
3. Letters: Assuming the password uses uppercase letters, there are 26 letters in total.

Let's analyze each part of the question:

Part (a)

How many different passwords can be formed?

1. Vowels: Choose 2 vowels out of 5, without repetition and with order considered. $\text{Choices for vowels} = 5 \times 4 = 20$

2. Even digits: Choose 2 even digits out of 5, without repetition. $\text{Choices for even digits} = 5 \times 4 = 20$

3. Letters: Choose 2 letters out of 26, without repetition. $\text{Choices for letters} = 26 \times 25 = 650$

4. Total passwords: $\text{Total} = 20 \times 20 \times 650 = 260,000$

Part (b)

How many different passwords can be formed if the password must contain the letter "J"?

• Since "J" is one of the two letters in the last part, we fix one letter as "J" and choose the second from the remaining 25 letters. $\text{Choices for letters with J} = 25$

• Thus, the total number of passwords becomes: $\text{Total with J} = 20 \times 20 \times 25 = 10,000$

Part (c)

How many different passwords can be formed if the password must contain the letter "A"?

• We follow a similar approach as in part (b), assuming "A" is one of the two letters. $\text{Choices for letters with A} = 25$

• Thus, the total number of passwords becomes: $\text{Total with A} = 20 \times 20 \times 25 = 10,000$

Part (d)

How many different passwords can be formed if the password must not contain the letters "O" or "P"?

• Since the restriction is on letters, there are only 24 choices for letters (excluding "O" and "P"). $\text{Choices for letters without O or P} = 24 \times 23 = 552$

• Thus, the total number of passwords becomes: $\text{Total without O or P} = 20 \times 20 \times 552 = 220,800$

Would you like more details on any part of these calculations?

Further Questions to Explore

1. How would the answer change if we allowed repetitions?
2. What if there was a constraint on the order of vowels and letters?
3. How many passwords would be possible if the structure were different, say, 3 letters and 3 digits?
4. What if we limited the letter choices to only consonants?
5. How would these calculations vary with lowercase letters included?

Tip

When dealing with combinatorics problems like this, breaking down each component and calculating step-by-step is essential for accuracy.