sonji bought a combination lock that opens with a four - digit number created using the digits 0 through 9…

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?\n84\n504\n3,024\n60,480
Answer
Explanation:
Step1: Determine available digits
Since the last - digit is 7 and digits can't be repeated, for the first digit of the remaining three, there are 9 available digits (0 - 9 excluding 7).
Step2: Determine second - remaining digit
After choosing the first digit, there are 8 remaining digits for the second digit of the remaining three.
Step3: Determine third - remaining digit
After choosing the first two digits, there are 7 remaining digits for the third digit of the remaining three.
Step4: Calculate the number of combinations
Using the permutation principle, the number of ways to choose the remaining 3 digits is (9\times8\times7=\frac{9!}{(9 - 3)!}=504).
Answer:
504