Let's solve the problem step by step:

We are tasked with finding the sum of the digits in the unit place of all the numbers formed using the digits $3, 4, 5, 6$, taken all at a time.

Step 1: Total number of permutations

Since there are 4 digits ($3, 4, 5, 6$) and we are forming numbers using all of them at a time, the total number of permutations is: $4! = 24$ So, 24 numbers will be formed.

Step 2: Contribution of each digit to the unit place

In each number, every digit appears in each position (unit place, tens place, etc.) an equal number of times. For 4 digits, each digit will appear in the unit place: $\frac{24}{4} = 6 \text{ times.}$

Step 3: Sum of the digits in the unit place

Each digit $3, 4, 5, 6$ contributes to the unit place 6 times. Therefore, the total contribution of all digits to the unit place is: $6 \times (3 + 4 + 5 + 6) = 6 \times 18 = 108$

Final Answer:

The sum of the digits in the unit place of all the numbers is $108$.

Correct Option:

(B) 108

Would you like further clarification on this problem or related concepts? Here are five questions for further exploration:

1. How can we generalize this method for $n$ digits?
2. What is the contribution of digits to other places (tens, hundreds, etc.) in the same problem?
3. How would the result change if the digits were repeated (e.g., $3, 3, 4, 5$)?
4. Can we derive the total sum of all the numbers formed in this way?
5. How does this problem relate to combinatorics principles in general?

Tip: When solving such problems, symmetry in permutations can simplify the computation for large datasets!