To determine how many four-digit numbers can be formed using the digits 2, 5, 6, 7, and 8, we'll need to consider the following:

1. Number of Digits Available: We have 5 digits (2, 5, 6, 7, 8).
2. Repetition Allowed or Not: Since the problem does not specify whether repetition of digits is allowed, we'll consider two cases: one where repetition is allowed and one where it is not.

Case 1: Repetition Allowed

• For each of the four positions in the number, we can choose any of the 5 digits.
• The total number of possible four-digit numbers is given by:

$5 \times 5 \times 5 \times 5 = 5^4 = 625$

Case 2: Repetition Not Allowed

• For the first digit, we have 5 choices.
• For the second digit, we have 4 choices (since one digit has already been used).
• For the third digit, we have 3 choices.
• For the fourth digit, we have 2 choices.

The total number of possible four-digit numbers is given by:

$5 \times 4 \times 3 \times 2 = 120$

Conclusion:

• With repetition allowed: 625 four-digit numbers can be formed.
• Without repetition: 120 four-digit numbers can be formed.

Would you like more details on any specific part, or have any questions?

Here are some related questions:

1. What is the number of three-digit numbers that can be formed using the digits 2, 5, 6, 7, and 8 without repetition?
2. How many different four-digit numbers can be formed using the digits 1, 3, 4, and 9 with repetition allowed?
3. If repetition is allowed, how many four-digit numbers can be formed using any digits from 0 to 9?
4. What is the probability of randomly selecting a four-digit number that is a multiple of 5 from the set of numbers formed with 2, 5, 6, 7, and 8?
5. How many even four-digit numbers can be formed using the digits 2, 5, 6, 7, and 8 without repetition?

Tip: Always check whether a problem allows repetition of digits; it significantly affects the total number of possible combinations.