To find how many 4-digit numbers can be formed under the given conditions, we must follow the rules:

1. The number must be divisible by 2, meaning the last digit must be even.
2. Digits cannot be repeated.

Step-by-Step Solution:

Step 1: Identify the possible digits for each position

• Thousands place (1st digit): The number cannot start with 0, so the possible choices for the first digit are from 1 to 9, giving us 9 choices.
• Units place (4th digit): Since the number must be divisible by 2, the last digit must be one of the even digits. The even digits available are {0, 2, 4, 6, 8}, giving us 5 choices.

Step 2: Consider the restrictions

Since the digits cannot be repeated, after selecting digits for the 1st and 4th positions, the choices for the 2nd and 3rd positions will decrease.

Step 3: Calculate the number of valid combinations

• Case 1: Last digit is 0
  If the last digit is 0, the remaining digits for the other positions can be chosen as follows:

  • For the thousands place, there are 9 options (1-9).
  • For the hundreds place, there are 8 options left.
  • For the tens place, there are 7 options left.

  So, the number of combinations when the last digit is 0 is: $9 \times 8 \times 7 = 504$

• Case 2: Last digit is 2, 4, 6, or 8
  If the last digit is 2, 4, 6, or 8, the thousands place can be chosen from 1 to 9, but one even digit is used up, so:

  • For the thousands place, there are still 8 choices (since one of the even digits is already used).
  • For the hundreds place, there are 8 options left.
  • For the tens place, there are 7 options left.

  The number of combinations for each of these cases is: $8 \times 8 \times 7 = 448$

Since there are 4 even digits to choose from (2, 4, 6, or 8), the total number of combinations for this case is: $448 \times 4 = 1792$

Step 4: Total number of combinations

Now, we can add the results of the two cases: $504 + 1792 = 2296$

Thus, 2296 different 4-digit numbers can be formed that are divisible by 2 and have no repeated digits.

---

Would you like more details on any step or have questions about this process?

Here are 5 related questions for deeper understanding:

1. How would the solution change if digits could be repeated?
2. What if the number had to be divisible by 5 instead of 2?
3. How many 3-digit numbers can be formed with the same restrictions?
4. How does allowing 0 in the thousands place affect the result?
5. What if the number must be divisible by 3?

Tip: When solving problems with digit restrictions, break the problem into smaller cases and count each case separately.