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The problem asks how many different 3-digit numbers can be made by arranging the 3 given number cards: 2, 4, and 5.

This is a problem of **permutations**. Since there are 3 distinct numbers, the number of different arrangements (or permutations) of these 3 digits is given by the factorial of the number of digits, which is \(3!\).

\[
3! = 3 \times 2 \times 1 = 6
\]

Thus, you can create **6 different 3-digit numbers** by arranging the digits 2, 4, and 5.

Here are the possible numbers:
- 245
- 254
- 425
- 452
- 524
- 542

Would you like more details or have any further questions?

Here are 5 related questions to consider:
1. What if one of the digits was repeated? How would that change the number of possible combinations?
2. How many different 4-digit numbers can be made from the digits 2, 4, 5, and 7?
3. How would the solution change if there were restrictions (e.g., even numbers only)?
4. Can this type of problem be generalized for any number of digits?
5. What if we wanted to count only the prime numbers that can be formed?

**Tip:** The number of arrangements of *n* distinct items is always calculated as \(n!\), where \(n!\) represents the factorial of \(n\).

How many different 3-digit numbers can be made by arranging the 3 number cards shown below (2, 4, and 5)?

Find out how to calculate the number of different 3-digit numbers that can be made by arranging the digits 2, 4, and 5. This math problem involves the concept of permutations and factorials, suitable for Grades 5-8.

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