Final answer:
The sum of the unit digits of all permutations of 3456 is calculated by multiplying the sum of the individual digits by the number of occurrences in the unit place. Each digit appears in the unit's place 6 times, leading to a total sum of 108. However, since the expected answer matches none of the options, the question might be flawed or misinterpreted.
Step-by-step explanation:
The question asks for the sum of the unit digits of all numbers formed with the digits 3, 4, 5, 6 taken at all times. This is a permutation problem, and since the numbers 3, 4, 5, 6 can each occupy the unit's place, when calculating the sum of the unit places for every permutation, each digit will appear an equal number of times in the unit place.
To find the total sum, first, determine the number of times each digit will appear in the unit place. There are 4! (which means 4 factorial) permutations possible, which is 4 x 3 x 2 x 1 = 24. Since there are 4 digits, each digit will appear 24/4 = 6 times in the unit place. So, the sum of unit digits for all numbers would be (3+4+5+6) x 6 = 108. However, only the units digit of this final sum is required, which is 8. Since 8 doesn't match any of the provided answer choices, it seems there has been a miscalculation or misinterpretation of the question in the original context.
Therefore, the sum of the digits in the unit place will be 3 + 4 + 5 + 6 = 18. Since we are considering the sum of digits, and 18 has two digits, we take the sum of the digits in 18, which is 1 + 8 = 9. So, the total sum of the digits in the unit place is 9 units.
However, we need to remember that in this case, the digit 0 is also considered. So, the final answer would be 9 + 0 = 9 units.