To determine how many numbers formed by using all the digits 1, 2, 3, 4, and 5 exactly once are even, follow these steps:

Step-by-Step Solution:

1. Identify the Requirement for the Number to be Even:

   • A number is even if its last digit is an even digit.
   • From the digits 1, 2, 3, 4, and 5, the even digits are 2 and 4.

2. Calculate the Number of Permutations for Each Even Digit as the Last Digit:

   • Case 1: Last digit is 2

     • Remaining digits: 1, 3, 4, 5
     • Number of permutations of the remaining 4 digits = $4!$
     • $4! = 4 \times 3 \times 2 \times 1 = 24$

   • Case 2: Last digit is 4

     • Remaining digits: 1, 2, 3, 5
     • Number of permutations of the remaining 4 digits = $4!$
     • $4! = 4 \times 3 \times 2 \times 1 = 24$

3. Add the Number of Permutations for Both Cases:

   • Total number of even numbers = $24 + 24 = 48$

Conclusion:

There are $48$ numbers that can be formed using the digits 1, 2, 3, 4, and 5 exactly once where the number is even.

Would you like more details or have any questions?

Here are 5 additional questions you might consider:

1. How many odd numbers can be formed using the digits 1, 2, 3, 4, and 5 exactly once?
2. How many numbers can be formed using the digits 1, 2, 3, 4, and 5 exactly once that are divisible by 5?
3. How many 4-digit numbers can be formed using the digits 1, 2, 3, and 4 exactly once that are divisible by 4?
4. How many numbers can be formed using the digits 1, 2, 3, and 4 exactly once that are multiples of 3?
5. How many 3-digit even numbers can be formed using the digits 1, 2, 3, and 4 exactly once?

Tip: When determining the number of possible permutations that satisfy certain conditions, consider fixing the digit(s) that meet the condition and then permuting the remaining digits.