Let's break this down step by step, using the Fundamental Counting Principle (FCP) and the Rule of Sum to solve this problem.

Key Points:

• We are asked for even numbers less than 500.
• The digits available are 1, 2, 3, 4, 5.
• Each digit is used only once in any given number.

Condition for Even Numbers:

• For a number to be even, its last digit must be 2 or 4, since those are the only even digits in our set (1, 2, 3, 4, 5).

Condition for Numbers Less than 500:

• The number must be less than 500, which means the first digit must be one of 1, 2, 3, or 4.

Case Breakdown:

We will consider different cases based on the number of digits (2-digit and 3-digit).

Case 1: 2-digit numbers

• The last digit must be 2 or 4 (even number condition).
• For each choice of the last digit, the first digit can be any of the remaining digits from the set, excluding the digit used for the last position.

Thus, we have:

1. Last digit = 2: The first digit can be 1, 3, 4, or 5 (4 choices).
2. Last digit = 4: The first digit can be 1, 2, 3, or 5 (4 choices).

So, the total number of 2-digit even numbers less than 500 is:

$4 \, (\text{choices for first digit when last digit is 2}) + 4 \, (\text{choices for first digit when last digit is 4}) = 8 \, \text{even numbers}$

Case 2: 3-digit numbers

• The last digit must be 2 or 4 (even number condition).
• The first digit must be one of 1, 2, 3, or 4 (since the number must be less than 500).
• The middle digit can be any of the remaining unused digits.

Let's break it down:

1. Last digit = 2: The first digit can be 1, 3, or 4 (3 choices, since 2 is already used). After choosing the first digit, we can select the middle digit from the remaining 3 digits (excluding the first and last digits).

   • For each choice of the first digit, there are 3 choices for the middle digit.
   • So, total for this case is: $3 \times 3 = 9$ 3-digit numbers.

2. Last digit = 4: The first digit can be 1, 2, or 3 (3 choices, since 4 is already used). Similarly, after choosing the first digit, there are 3 remaining choices for the middle digit.

   • For each choice of the first digit, there are 3 choices for the middle digit.
   • So, total for this case is: $3 \times 3 = 9$ 3-digit numbers.

Thus, the total number of 3-digit even numbers less than 500 is:

$9 \, (\text{choices for first and middle digits when last digit is 2}) + 9 \, (\text{choices for first and middle digits when last digit is 4}) = 18 \, \text{even numbers}$

Final Total:

Now, let's add the numbers from both cases:

$8 \, (\text{2-digit numbers}) + 18 \, (\text{3-digit numbers}) = 26 \, \text{even numbers less than 500}$

Conclusion:

There are 26 even numbers less than 500 that can be made from the digits 1, 2, 3, 4, and 5, with each digit being used only once.

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Would you like further clarification on any of the steps, or do you have another question?

Here are a few related questions for you to explore further:

1. How would the count change if the number was restricted to only 2 digits?
2. What if the number could use repeated digits, how would that affect the total?
3. How many odd numbers less than 500 can be formed with the same set of digits?
4. What would happen if the number was not restricted to be less than 500?
5. Can you solve a similar problem where the number must be divisible by 3?

Tip: When solving counting problems, always carefully break down the cases based on constraints like "even" or "less than 500" before applying the counting principles.