Let f(n) denotes the number of different ways, the positive integer n ca be expressed as the sum of the 1's and 2's. for example, f(4)=5.
i.e., `4=1+1+1+1`
`=1+1+2=1+2+1=2+1+1=2+2`
Q. In a stage show, f(4) superstars and f(3) junior artists participate. each one is going to present one item, then the number of ways the sequence of items can be planned, if no two junior artists present their items consecutively, is

Options

A. 144
B. 360
C. 4320
D. 14400

Explanation

To solve the problem, we need to find the number of ways to arrange f(4) superstars and f(3) junior artists such that no two junior artists perform consecutively.

### Step-by-Step Solution:

1. **Identify f(4) and f(3)**:
- We know from the problem that \( f(4) = 5 \) (the number of ways to express 4 as sums of 1's and 2's).
- We need to calculate \( f(3) \):
- \( f(3) \) can be expressed as:
- \( 1 + 1 + 1 \)
- \( 1 + 2 \)
- \( 2 + 1 \)
- Thus, \( f(3) = 3 \).

2. **Determine the number of superstars and junior artists**:
- Number of superstars = \( f(4) = 5 \)
- Number of junior artists = \( f(3) = 3 \)

3. **Arrange the superstars**:
- First, we arrange the 5 superstars. The arrangement can be done in \( 5! \) ways.
- \( 5! = 120 \).

4. **Identify positions for junior artists**:
- When the 5 superstars are arranged, they create gaps for the junior artists.
- The arrangement of 5 superstars creates 6 gaps (one before each superstar and one after the last superstar):
- _ S _ S _ S _ S _ S _
- Therefore, we have 6 positions to place the junior artists.

5. **Choose positions for junior artists**:
- We need to select 3 out of these 6 positions for the junior artists. This can be done in \( \binom{6}{3} \) ways.
- \( \binom{6}{3} = 20 \).

6. **Arrange the junior artists**:
- The 3 junior artists can be arranged among themselves in \( 3! \) ways.
- \( 3! = 6 \).

7. **Calculate the total arrangements**:
- The total number of arrangements is given by the product of the arrangements of superstars, the selection of positions for junior artists, and the arrangements of junior artists:
\[ \text{Total arrangements} = \binom{6}{3} \times 5! \times 3! = 20 \times 120 \times 6 \]

8. **Perform the final calculation**:
- Calculate \( 20 \times 120 = 2400 \)
- Then, \( 2400 \times 6 = 14400 \).

### Final Answer:
The total number of ways the sequence of items can be planned is **14400**.