Final answer:
The sum of all four-digit numbers formed using the digits 1, 2, 4, and 6 is arrived at by calculating the sum of permutations for each digit across all place values, weighted accordingly, resulting in a final sum of 88,858.
Step-by-step explanation:
The student is asking to calculate the sum of all four-digit numbers that can be formed using the digits 1, 2, 4, and 6. Each digit can appear only once in each number. To solve this, we recognize that each digit will appear an equal number of times in each place value (thousands, hundreds, tens, and ones).
Since there are 4 digits, and we want to form 4-digit numbers, we can use permutations to calculate that there are 4! (which means 4 factorial) different numbers we can create, 4! = 4 × 3 × 2 × 1 = 24 numbers. The sum for each place value will be the sum of the digits (1+2+4+6) multiplied by 6 (since each digit appears in each place value 6 times).
This gives us a sum of 13 × 6 for each place value, which is 78 for each place. We weight this by the place value (for thousands 1000, for hundreds 100, and so on) and then sum those up to find the total sum of all numbers. The calculation should look like this:
- (78 × 1000) for the thousands place
- (78 × 100) for the hundreds place
- (78 × 10) for the tens place
- (78 × 1) for the ones place
Adding these together:
78000 + 7800 + 780 + 78 = 88858
The correct option is 2. 88,858.