The problem asks how many different possible combinations there are for a locker with 10 numbered dials, where each dial can use the digits 0 through 9, and each digit can only be used once.

Step-by-Step Solution:

1. Total Digits Available: The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, which gives a total of 10 digits.

2. Combinations Without Repetition: Since each digit can only be used once and there are 10 dials, this becomes a permutation problem where we are choosing and arranging 10 digits out of the 10 available.

   The number of different ways to arrange 10 digits is given by: $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$

3. Conclusion: The total number of different possible combinations is $3,628,800$, which is much larger than any of the options listed.

Given the current options, none of the choices correspond to the correct value of $10!$. This seems to be a mismatch, as the question setup points to a problem involving permutations, but no answer close to 3.6 million is provided.

Do you have any other questions about the reasoning, or would you like further details?

Related Questions:

1. What is the difference between permutations and combinations?
2. How would the answer change if repetition of digits were allowed?
3. How can you quickly calculate large factorials, such as 10!?
4. How many combinations are there if we only use 4 digits out of 10?
5. How does using a smaller set of digits (e.g., 5 out of 10) affect the total number of possible permutations?

Tip: When faced with a problem involving "each digit can only be used once," think of permutations rather than combinations!