You're asking for all possible combinations of three numbers chosen from 0 to 9. Here are the key results and how to interpret them. Direct answer
- If you mean all ordered three-digit strings where each position can be 0–9 (i.e., 000 to 999), there are 10 × 10 × 10 = 1000 such combinations.
- If you mean all three-digit numbers with possible repetition of digits allowed and order matters, it’s still 1000 possibilities (000–999).
- If you mean all unique-digit three-number sequences (no repetition of digits and order matters), there are 10 choices for the first digit, 9 for the second, and 8 for the third, giving 10 × 9 × 8 = 720 possibilities.
- If you mean all combinations of three digits chosen from {0,…,9} where the order does not matter (i.e., sets of three digits without regard to order), the count is 10 choose 3 = 120 possibilities. Note that this last interpretation ignores the arrangement of digits.
Clarifications and notes
- For a standard 3-wheel lock with digits 0–9 on each wheel, each wheel is independent, so the total is 10^3 = 1000, covering 000 through 999 in order. If you’re enumerating, you’d list every triple (d1, d2, d3) with di ∈ {0,…,9}.
- If you require exactly three distinct digits (no repeats) and care about the order, use 720 combinations. If repeats are allowed but you don’t care about order, that’s a different count (for example, combinations with repetition: C(10+3−1,3) = C(12,3) = 220, which is a different interpretation and depends on what “combinations” means in your context).
If you’d like, specify which interpretation you want (ordered with repeats, ordered without repeats, unordered without repeats, unordered with repeats), and I’ll provide the exact counts and a listing method or sample for that case.
