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 `B_3` can have only `C_1` or `C_5`, in how many ways can you arrange the caps such that all boxes have one cap?

Options

A. 480
B. 420
C. 48
D. 88

Answer

To solve the problem, we need to determine how many ways we can arrange 5 different caps into 5 different boxes, given that box B3 can only contain either cap C1 or cap C5.

### Step-by-Step Solution:

1. **Identify the Options for Box B3**:
- Box B3 can only have either cap C1 or cap C5. Therefore, we have 2 choices for box B3.

2. **Select a Cap for Box B3**:
- Let's consider the two cases separately:
- Case 1: B3 contains C1.
- Case 2: B3 contains C5.

3. **Remaining Caps After Filling Box B3**:
- After choosing a cap for B3, we will have 4 caps left for the other boxes.
- In both cases (whether we chose C1 or C5 for B3), the remaining caps will be the other 4 caps.

4. **Arranging the Remaining Caps in the Other Boxes**:
- The remaining caps can be arranged in the remaining boxes (B1, B2, B4, B5).
- Since all boxes and caps are unique, we can use permutations to arrange the remaining 4 caps.
- The number of ways to arrange 4 unique caps in 4 unique boxes is given by \(4!\) (4 factorial).

5. **Calculate the Total Arrangements**:
- The total arrangements can be calculated as:
\[ \text{Total Arrangements} = (\text{Choices for B3}) \times (\text{Arrangements of remaining caps}) \]
- This gives us:
\[ \text{Total Arrangements} = 2 \times 4! = 2 \times 24 = 48 \]

### Final Answer:
The total number of ways to arrange the caps such that all boxes have one cap is **48**.