To solve this problem, let's break it down step by step.

Step 1: Identify whether it is a Permutation or a Combination problem.

• A combination is when the order of items doesn't matter (e.g., selecting lottery numbers).
• A permutation is when the order of items does matter (e.g., arranging numbers in a specific order).

Since the digits are being arranged in a lock where the order matters (e.g., 123 is different from 321), this is a permutation problem, not a combination.

Step 2: Determine the number of possible arrangements.

• The lock has 3 digits, and each digit can be chosen from 0 to 9 (which gives 10 digits total: 0, 1, 2, ..., 9).
• However, the problem states that no digit can be used twice.

So, we calculate the number of arrangements as a permutation:

• For the first digit, there are 10 options (since any digit from 0 to 9 can be chosen).
• For the second digit, there are 9 options left (since one digit has already been used).
• For the third digit, there are 8 options left (since two digits have already been used).

Thus, the total number of permutations is: $10 \times 9 \times 8 = 720$

Step 3: Conclusion

The correct answer is 720 permutations, since the order of digits matters and no digits are repeated.

Correct Answer:

C. 720; Permutation

Would you like more details or have any questions about this process?

Here are five related questions to expand your understanding:

1. How do you distinguish between a permutation and a combination in a problem?
2. What happens to the number of permutations if repetition of digits were allowed?
3. How would the solution change if the lock had 4 digits instead of 3?
4. What is the formula for calculating permutations in general?
5. How can you use factorial notation to simplify solving permutation problems?

Tip: In any permutation problem, remember that the order of selection matters!