Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?

Options

A. 8
B. 10
C. 15
D. 22

Answer

To solve the problem of arranging boxes numbered 1 to 5 with either a red or a blue ball, such that no two adjacent boxes contain blue balls, we can follow these steps:

### Step 1: Understand the Total Arrangements
First, we need to determine the total number of arrangements without any restrictions. Since each box can either contain a red ball (R) or a blue ball (B), and there are 5 boxes, the total arrangements can be calculated as:
\[ \text{Total arrangements} = 2^5 = 32 \]

**Hint:** Remember that each box has 2 choices (red or blue), and you can use the power of 2 for the total number of boxes.

### Step 2: Identify Invalid Arrangements
Next, we need to find the arrangements that violate the condition of no two adjacent blue balls. We will consider cases where blue balls are adjacent.

1. **Case 1: 2 adjacent blue balls**
The pairs of adjacent boxes that can contain blue balls are:
- (1, 2)
- (2, 3)
- (3, 4)
- (4, 5)
This gives us 4 ways to choose the position of the adjacent blue balls. After placing the blue balls, we have 3 remaining boxes which can be filled with either red or blue balls (but no additional blue balls can be placed adjacent to the existing ones). Thus, each of the 3 remaining boxes can be filled with red or blue (but not adjacent to blue), giving us:
\[ \text{Ways} = 4 \times 2^3 = 4 \times 8 = 32 \]

2. **Case 2: 3 adjacent blue balls**
The groups of 3 adjacent boxes can be:
- (1, 2, 3)
- (2, 3, 4)
- (3, 4, 5)
This gives us 3 ways to choose the position of the 3 adjacent blue balls. After placing these, we have 2 remaining boxes, which can be filled in:
\[ \text{Ways} = 3 \times 2^2 = 3 \times 4 = 12 \]

3. **Case 3: 4 adjacent blue balls**
The groups of 4 adjacent boxes can be:
- (1, 2, 3, 4)
- (2, 3, 4, 5)
This gives us 2 ways to choose the position of the 4 adjacent blue balls. After placing these, we have 1 remaining box:
\[ \text{Ways} = 2 \times 2^1 = 2 \times 2 = 4 \]

4. **Case 4: 5 blue balls**
There is only 1 arrangement where all boxes are filled with blue balls:
\[ \text{Ways} = 1 \]

### Step 3: Calculate Total Invalid Arrangements
Now, we sum all the invalid arrangements:
\[ \text{Total invalid arrangements} = 32 + 12 + 4 + 1 = 49 \]

### Step 4: Calculate Valid Arrangements
Finally, we subtract the total invalid arrangements from the total arrangements:
\[ \text{Valid arrangements} = \text{Total arrangements} - \text{Total invalid arrangements} = 32 - 49 = -17 \]

However, this result indicates an error in our counting, as the valid arrangements cannot be negative. Let's correct our approach:

Instead, we can use the Fibonacci sequence approach, where:
- Let \( a_n \) be the number of valid arrangements for \( n \) boxes.
- The recurrence relation is \( a_n = a_{n-1} + a_{n-2} \) with base cases \( a_1 = 2 \) (R, B) and \( a_2 = 3 \) (RR, RB, BR).

Calculating:
- \( a_3 = a_2 + a_1 = 3 + 2 = 5 \)
- \( a_4 = a_3 + a_2 = 5 + 3 = 8 \)
- \( a_5 = a_4 + a_3 = 8 + 5 = 13 \)

Thus, the total valid arrangements for 5 boxes is **13**.

### Final Answer
The number of different arrangements possible is **13**.