Question

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

Explanation:

Step1: Identify available digits

We have digits 0-9, minus 7 (used as last digit), so 9 remaining digits. We need to choose 3 distinct digits for the first three positions, where order matters.

Step2: Calculate permutations

For the first position: 9 choices.
For the second position: 8 remaining choices.
For the third position: 7 remaining choices.
Total permutations = $9 \times 8 \times 7$

Step3: Compute the product

$9 \times 8 = 72$; $72 \times 7 = 504$

Answer:

504 (Option B)