Let's walk through how to calculate each of these cases to verify the results:

Part 1: No digit is repeated

If no digit is repeated, we are selecting 5 digits out of 10. This is a permutation problem, where the order of the digits matters.

The formula for permutations when choosing $r$ elements from $n$ total elements is:

$P(n, r) = \frac{n!}{(n - r)!}$

In this case:

• $n = 10$ (the number of digits: 0-9)
• $r = 5$ (we are selecting 5 digits)

The number of possible combinations is:

$P(10, 5) = \frac{10!}{(10-5)!} = \frac{10!}{5!}$

Calculating $P(10, 5)$:

$P(10, 5) = 10 \times 9 \times 8 \times 7 \times 6 = 30,240$

So, 30,240 is correct for this case.

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Part 2: Digits can be repeated

If digits can be repeated, then each of the 5 wheels has 10 possible digits (0-9).

Since each position can be any of the 10 digits, the total number of combinations is:

$10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100,000$

So, 100,000 is also correct for this case.

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Part 3: Successive digits must be different

If successive digits must be different, we are dealing with a slightly modified scenario. The first digit can be any of the 10 digits, but for each subsequent digit, it must be different from the previous one.

• For the first digit, there are 10 possible choices.
• For the second digit, there are 9 possible choices (it must be different from the first).
• For the third digit, there are again 9 choices (it must be different from the second but not necessarily from the first).
• For the fourth digit, there are 9 choices (different from the third).
• For the fifth digit, there are 9 choices (different from the fourth).

Thus, the total number of combinations is:

$10 \times 9 \times 9 \times 9 \times 9 = 10 \times 9^4$

Calculating $9^4$:

$9^4 = 6561$

So, the total number of combinations is:

$10 \times 6561 = 65,610$

It seems like the value you provided (7,290) doesn't match this calculation. The correct number should be 65,610, not 7,290.

Would you like further details or clarification?

Follow-up questions:

1. How would the calculations change if the lock had 6 wheels instead of 5?
2. What if we were allowed to use only even digits for the combinations?
3. How does the approach change if we must avoid two or more consecutive digits being the same?
4. What is the probability of guessing the correct combination if digits can be repeated?
5. How many combinations are possible if no two consecutive digits can differ by more than 1?

Tip:

When dealing with combinations and permutations, always verify if the order of the selection matters, as that dictates whether to use combinations or permutations.