1. **Stating the problem:** We need to form a 5-digit number using digits from 0 to 9 without repetition. The number cannot start with zero and must be even. 2. **Understanding the conditions:** - The first digit cannot be 0. - The last digit must be even (0, 2, 4, 6, 8). - No digit repeats. 3. **Step 1: Choose the last digit (even digit):** Possible even digits are 0, 2, 4, 6, 8, so 5 choices. 4. **Step 2: Choose the first digit:** The first digit cannot be zero and cannot be the last digit chosen. - If last digit is 0, first digit choices are from {1-9} = 9 choices. - If last digit is not 0, first digit choices are from {1-9} excluding last digit = 8 choices. 5. **Step 3: Choose the middle three digits:** After choosing first and last digits, 8 digits remain. We need to choose 3 digits without repetition and order matters. Number of ways = permutations of 8 digits taken 3 at a time = $P(8,3) = 8 \times 7 \times 6 = 336$. 6. **Calculate total numbers:** - Case 1: Last digit = 0 Number of numbers = choices for last digit (1) \times choices for first digit (9) \times choices for middle three digits (336) = $1 \times 9 \times 336 = 3024$ - Case 2: Last digit is one of {2,4,6,8} (4 choices) Number of numbers = choices for last digit (4) \times choices for first digit (8) \times choices for middle three digits (336) = $4 \times 8 \times 336 = 10752$ 7. **Total number of such 5-digit numbers:** $$3024 + 10752 = 13776$$ **Final answer:** There are $13776$ such 5-digit numbers.