1. **Problem:** How many different 4-digit even numbers can be formed from the digits 1, 3, 5, 6, 8, and 9 if no repetition of digits is allowed? 2. **Formula and rules:** - A 4-digit even number must end with an even digit. - The digits available are 1, 3, 5, 6, 8, 9. - Even digits here are 6 and 8. - No repetition means each digit can be used only once. 3. **Step-by-step solution:** 1. Choose the last digit (units place) which must be even: 2 choices (6 or 8). 2. Choose the first digit (thousands place): cannot be zero and cannot be the digit chosen for the last place. So from remaining 5 digits, 5 choices. 3. Choose the second digit (hundreds place): from remaining 4 digits, 4 choices. 4. Choose the third digit (tens place): from remaining 3 digits, 3 choices. 4. **Calculate total number of such numbers:** $$\text{Total} = 2 \times 5 \times 4 \times 3 = 120$$ 5. **Explanation:** - We first fix the last digit to ensure the number is even. - Then we select the first digit from the remaining digits (excluding the last digit). - Then fill the middle two digits from the remaining digits without repetition. **Final answer:** There are **120** different 4-digit even numbers possible.