Sonji bought a combination lock that opens with a four-digit number created using the digits 0 through 9 . The same digit cannot be used more than once in the combination. If Sonji wants the last digit to be a 7 and the order of the digits matters, how many ways can the remaining digits be chosen? 84 504 3,024 60,480

To find the number of ways to choose the remaining digits, let’s break it down step by step. Since the last digit is fixed as 7, we have three digits left to choose from the digits 0-9, excluding 7. This leaves us with 9 possible digits: 0, 1, 2, 3, 4, 5, 6, 8, and 9. 1. For the first digit, we can choose any of the 9 digits. 2. For the second digit, we can choose from the remaining 8 (since one digit has been used). 3. For the third digit, we can choose from the remaining 7. So, the total number of combinations is given by: \[ 9 \times 8 \times 7 = 504. \] Therefore, the answer is **504**.