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

1 answer

To determine how many ways Sonji can choose the digits for her combination lock given that the last digit must be 7 and the same digit cannot be used more than once, we can follow these steps:

  1. Fix the Last Digit: Since the last digit is already fixed as 7, we only need to choose the first three digits from the remaining digits, which are 0-6 and 8-9 (a total of 9 available digits).

  2. Choosing the First Digit: For the first digit, Sonji can choose any of the 9 remaining digits.

  3. Choosing the Second Digit: Once the first digit is chosen, that digit cannot be reused. Therefore, for the second digit, she has 8 choices left.

  4. Choosing the Third Digit: After choosing the first two digits, there are now 7 remaining digits to choose from for the third digit.

Now we can calculate the total number of combinations:

\[ \text{Total Combinations} = (\text{choices for first digit}) \times (\text{choices for second digit}) \times (\text{choices for third digit}) \]

\[ \text{Total Combinations} = 9 \times 8 \times 7 \]

Calculating this step-by-step:

  • First, calculate \(9 \times 8 = 72\).
  • Then, calculate \(72 \times 7 = 504\).

Thus, the total number of ways Sonji can choose the remaining digits for her combination lock, while keeping the last digit as 7, is \(504\).

So, the answer is \(\boxed{504}\).