Analyze the scenario where a password consists of 3 letters followed by 3 digits; how many distinct passwords can be created if letters can repeat but digits cannot?
user
Yes, sure. So, in this scenario, you have 36 (26 letters + 10 digits) units to choose from, but since the digits are not allowed to repeat, you cannot use all of them. In that case, 10 digits can be arranged in 3 positions, and thus the total number of distinct passwords would be 10 * 3! = 600.

Now, if the letters were allowed to repeat, then their position in the password would not matter or count since they can be interchanged. Hence, the total number of distinct passwords would be 26 * 3 = 78, but since we're excluding cases like repeated digits, we do not consider this.

So, the final answer to your question in this case would be 600, excluding the cases where digits are repeated. I hope this clears it for you.
assistant
Analyze the scenario where a password consists of 3 letters followed by 3 digits; how many distinct passwords can be created if letters can repeat but digits cannot?
user
Sure, let's break this down.

The password consists of two parts: a 3-letter part and a 3-digit part.

1. For the 3-letter part, since the letters can repeat, we have 26 choices for each letter (assuming we're using the English alphabet). So, the total number of ways to choose the 3 letters is 26 * 26 * 26 = 17,576.

2. For the 3-digit part, since the digits cannot repeat, we have 10 choices for the first digit (0-9), 9 choices for the second digit (since one digit has been used), and 8 choices for the third digit. So, the total number of ways to choose the 3 digits is 10 * 9 * 8 = 720.

Since these two parts are independent, we multiply the number of ways for each part to get the total number of distinct passwords.

So, the total number of distinct passwords is 17,576 * 720 = 12,691,200.
assistant