A locker combination system uses four digits from 0 to 9. How many different four-digit combinations are possible if the first and the last digit cannot be zero and no digit can be repeated
The correct answer and explanation is:
The problem asks how many different four-digit combinations can be made with the digits 0 to 9, under the condition that the first and last digits cannot be zero, and no digit can be repeated.
To solve this, let’s break it down step by step:
Step 1: First digit
The first digit cannot be zero, so it can be any digit from 1 to 9. This gives us 9 possible choices for the first digit.
Step 2: Second digit
The second digit can be any digit from 0 to 9, except the first digit (since no digit can be repeated). Therefore, after selecting the first digit, there are 9 remaining digits to choose from.
Step 3: Third digit
The third digit can also be any digit from 0 to 9, except the first and second digits. This gives 8 remaining choices for the third digit.
Step 4: Last digit
The last digit cannot be zero, and it must also be different from the first, second, and third digits. This gives 7 remaining choices for the last digit.
Total combinations
To find the total number of different combinations, we multiply the number of choices for each digit: 9×9×8×7=45369 \times 9 \times 8 \times 7 = 4536
Thus, the total number of possible four-digit combinations is 4536.
Explanation:
The process of determining the number of different combinations relies on understanding the constraints of the problem:
- The first and last digits cannot be zero.
- Digits cannot repeat.
By applying these restrictions and systematically counting the available choices for each digit position, we arrive at the final answer of 4536.