There are 5 different caps `c_1,c_2,c_3,c_4` and `c_5` and 5 different boxes `B_1,B_2,B_3,B_4` and `B_5`. The capacity of each box is sufficient to accommodate all the 5 caps.
If all the caps are identical, in how many ways can the caps be arranged in the different boxes such that no box is empty?

Options

A. 1
B. 2
C. 6
D. 8

Answer

To solve the problem of distributing 5 identical caps into 5 different boxes such that no box is empty, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 5 identical caps and 5 different boxes. Since the caps are identical, the arrangement of caps in boxes will be based solely on how many caps are placed in each box, not on which specific cap goes into which box.

2. **Condition of Non-emptiness**: Since no box can be empty, we must place at least one cap in each box. Given that we have 5 caps and 5 boxes, the only way to satisfy this condition is to place exactly one cap in each box.

3. **Distribution of Caps**: Since all caps are identical, placing one cap in each of the 5 boxes is the only arrangement possible. There are no other combinations since if we were to place more than one cap in any box, at least one box would end up empty, violating the condition.

4. **Counting the Arrangements**: As all the caps are identical, the arrangement of caps in boxes does not change the outcome. Therefore, there is only one way to distribute the caps such that no box is empty.

5. **Final Answer**: The total number of ways to arrange the caps in the boxes, ensuring no box is empty, is **1**.

### Conclusion:
Thus, the answer to the problem is **1**.