Question

a computer password is to consist of six letters, none of which are used more than once. For security reasons, the letters Q and Z are not to be used in the first position how many different passwords are possible?

Answer 1

To find the total number of different passwords that can be formed consisting of six letters (with the condition that no letter is used more than once and that the letters Q and Z cannot be used in the first position), we can follow these steps:

1. **Determine the allowed letters for the first position:**
- The English alphabet has 26 letters.
- Removing Q and Z leaves us with \( 26 - 2 = 24 \) possible letters for the first position.

2. **Choose the letters for the remaining positions:**
- After the first letter has been chosen, we can use all remaining letters for the subsequent positions:
- For the second position, we have 25 letters left (since we cannot use the letter chosen for the first position).
- For the third position, we have 24 letters left.
- For the fourth position, we have 23 letters left.
- For the fifth position, we have 22 letters left.
- For the sixth position, we have 21 letters left.

3. **Calculate the total number of combinations:**
- The total number of different passwords can then be calculated by multiplying the number of choices for each position:

\[
\text{Total passwords} = \text{Choices for first position} \times \text{Choices for second position} \times \text{Choices for third position} \times \text{Choices for fourth position} \times \text{Choices for fifth position} \times \text{Choices for sixth position}
\]

Substituting the number of choices we have:

\[
\text{Total passwords} = 24 \times 25 \times 24 \times 23 \times 22 \times 21
\]

Now let's perform this multiplication step-by-step:

1. \( 24 \times 25 = 600 \)
2. \( 600 \times 24 = 14400 \)
3. \( 14400 \times 23 = 331200 \)
4. \( 331200 \times 22 = 7298400 \)
5. \( 7298400 \times 21 = 153722880 \)

Thus, the total number of different possible passwords is:

\[
\boxed{153722880}
\]