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

Answer 1

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 \(\boxed{504}\).