Final answer:
To find the sum of all 4-digit numbers formed from the digits 1, 2, 3, and 4 without repetition, each digit appears in each position 6 times across the 24 unique combinations, resulting in a total sum of 66,660.
Step-by-step explanation:
The question asks for the sum of all 4-digit numbers that can be formed from the digits 1, 2, 3, and 4, without repetition of digits. To calculate the sum, we first need to understand that the number of combinations of 4 digits is given by 4 factorial (4!), which is 4×3×2×1, resulting in 24 different combinations. Each digit will appear an equal number of times in each position.
Since there are 6 combinations where each digit can be in each place (24/4), we can determine the sum of each position separately. The sum of each digit in each position is 6*(1+2+3+4) = 60. Therefore, the sum of the thousands place is 60,000, hundreds place 6,000, tens place 600, and ones place 60. Adding these sums together gives us 60,000 + 6,000 + 600 + 60 = 66,660.