To determine the number of possible combinations for the 4-digit bike lock, where each space can hold a number between 0 and 9 and no digit can be repeated, we can use the following steps:

1. First digit: There are 10 possible digits (0 through 9) to choose from.
2. Second digit: Since the first digit is already used, we have 9 remaining digits to choose from.
3. Third digit: After choosing two digits, we have 8 digits left to pick from.
4. Fourth digit: Finally, we have 7 digits remaining to choose for the last spot.

Thus, the total number of possible combinations is:

$10 \times 9 \times 8 \times 7 = 5040$

So, there are 5040 possible combinations for the bike lock.

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. What if two digits could be the same? How many combinations would that give?
2. How does the number of combinations change if the lock had 5 digits instead of 4?
3. What if the lock allowed only even digits (0, 2, 4, 6, 8)? How many combinations could be made?
4. How would you calculate combinations if the numbers must increase from left to right?
5. How would adding one additional restriction, such as the first digit cannot be 0, affect the total?

Tip: When working with restrictions in counting problems, consider using permutations (for ordered arrangements) or combinations (for unordered selections).