Craig invited four friends to watch a TV show. He arranged 5 seats in a row. The number of ways he and his four friends can sit in the row is n. In how many of these ways can Craig sit in the middle?

Options

A. n
B. `n/2`
C. `n/3`
D. `n/5`

Answer

To solve the problem, we need to determine how many ways Craig and his four friends can sit in a row of five seats, with the condition that Craig must sit in the middle seat.

### Step-by-Step Solution:

1. **Total Arrangement of Friends**:
- There are 5 seats and 5 people (Craig + 4 friends).
- The total number of ways to arrange 5 people in 5 seats is given by \(5!\) (5 factorial).
- Calculation:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
- Therefore, the total number of arrangements \(n = 120\).

2. **Craig Sitting in the Middle**:
- If Craig is to sit in the middle seat, we fix Craig in the middle position.
- This leaves us with 4 seats to fill with his 4 friends.

3. **Arrangement of Friends**:
- The number of ways to arrange the 4 friends in the remaining 4 seats is given by \(4!\) (4 factorial).
- Calculation:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

4. **Expressing in Terms of n**:
- We know \(n = 120\).
- We need to express the number of ways Craig can sit in the middle in terms of \(n\).
- Since \(24\) (the number of arrangements with Craig in the middle) can be expressed as:
\[ 24 = \frac{120}{5} = \frac{n}{5} \]

5. **Final Answer**:
- Therefore, the number of ways in which Craig can sit in the middle is \(\frac{n}{5}\).

### Conclusion:
The final answer is \(\frac{n}{5}\).

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