The sum of the digits in unit place of all the numbers formed using the digits 3,4,5,6 without repetitions, the no. of such numbers are

Options

A. A. 432
B. B. 108
C. C. 36
D. D. 18

Answer

To solve the problem of finding the sum of the digits in the unit place of all the numbers formed using the digits 3, 4, 5, and 6 without repetitions, we can follow these steps:

### Step 1: Determine the total number of permutations
We have 4 digits: 3, 4, 5, and 6. Since we are forming 4-digit numbers using all these digits without repetition, the total number of permutations can be calculated as:

\[ \text{Total permutations} = 4! = 24 \]

### Step 2: Identify the unit place contributions
Each digit (3, 4, 5, 6) will appear in the unit place of the numbers formed. Since there are 24 total permutations and we have 4 digits, each digit will appear in the unit place an equal number of times.

\[ \text{Number of times each digit appears in the unit place} = \frac{24}{4} = 6 \]

### Step 3: Calculate the contribution of each digit in the unit place
Now we can calculate the contribution of each digit when it appears in the unit place:

- Contribution of 3 in unit place: \(3 \times 6 = 18\)
- Contribution of 4 in unit place: \(4 \times 6 = 24\)
- Contribution of 5 in unit place: \(5 \times 6 = 30\)
- Contribution of 6 in unit place: \(6 \times 6 = 36\)

### Step 4: Sum the contributions
Now, we add up all the contributions from each digit:

\[ \text{Total sum} = 18 + 24 + 30 + 36 \]

Calculating this gives:

\[ \text{Total sum} = 108 \]

### Conclusion
The sum of the digits in the unit place of all the numbers formed using the digits 3, 4, 5, and 6 without repetitions is **108**.

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