1. **Problem statement:** Find the number of 4-digit numbers formed from digits 0,1,2,3,4,5,6,7 such that each number contains the digit 1 at least once. 2. **Total 4-digit numbers from given digits:** - The first digit can be from 1 to 7 (cannot be 0 as it won't be a 4-digit number), so 7 choices. - Each of the remaining 3 digits can be any of the 8 digits (0 to 7), so $8^3 = 512$ choices. - Total numbers = $7 \times 512 = 3584$. 3. **Numbers without digit 1:** - Digits allowed: 0,2,3,4,5,6,7 (7 digits, excluding 1). - First digit cannot be 0, so 6 choices (2,3,4,5,6,7). - Each of the remaining 3 digits can be any of the 7 digits. - Total numbers without 1 = $6 \times 7^3 = 6 \times 343 = 2058$. 4. **Numbers with at least one digit 1:** - Use complementary counting: $$\text{Numbers with 1} = \text{Total} - \text{Without 1} = 3584 - 2058 = 1526.$$ 5. **Check options:** None of the options exactly match 1526, but closest is (c) 1522, which might be a typo or approximation. **Final answer:** 1526 (closest option is (c) 1522).