There are 7 seats in a row. Three persons take seats at random the probability that the middle seat is always occupied and no two persons are consecutive is

Options

A. `9//70`
B. `9//35`
C. `4//35`
D. none of these

Answer

To solve the problem step by step, we will analyze the conditions given and calculate the required probability.

### Step 1: Understand the seating arrangement
We have 7 seats in a row, labeled as:
1, 2, 3, 4, 5, 6, 7

The middle seat is seat number 4.

### Step 2: Fix the middle seat
Since the middle seat (seat 4) must always be occupied, we can place one person in seat 4. This leaves us with 6 remaining seats (1, 2, 3, 5, 6, 7) for the other two persons.

### Step 3: Ensure no two persons are consecutive
To ensure that no two persons are sitting consecutively, we will analyze the remaining seats. After placing one person in seat 4, we need to place the other two persons in such a way that they are not next to each other.

The available seats are now:
1, 2, 3, 5, 6, 7

### Step 4: Identify valid seat combinations
We can visualize the available seats as:
- If we place a person in seat 1, the next person can occupy either seat 3, 5, 6, or 7.
- If we place a person in seat 2, the next person can occupy either seat 5, 6, or 7.
- If we place a person in seat 3, the next person can occupy either seat 5, 6, or 7.
- If we place a person in seat 5, the next person can occupy either seat 1, 2, or 3.
- If we place a person in seat 6, the next person can occupy either seat 1, 2, or 3.
- If we place a person in seat 7, the next person can occupy either seat 1, 2, or 3.

To ensure that no two persons are consecutive, we can select the two remaining persons from the available seats (1, 2, 3, 5, 6, 7) while maintaining the condition of non-consecutiveness.

### Step 5: Calculate favorable outcomes
To find the number of ways to select 2 seats from the 6 available seats (1, 2, 3, 5, 6, 7) such that no two selected seats are consecutive, we can use the following reasoning:

1. **Choose 2 seats from 6 available seats**:
- If we denote the selected seats as X, we can represent the arrangement as follows:
- X _ X _ _ _ (where _ represents the remaining seats)
- The gaps between the selected seats must be at least one seat apart.

2. **Using the gap method**:
- We can think of placing "gaps" between the selected seats. If we choose 2 seats, we will have 4 remaining seats to fill the gaps.
- The arrangement can be visualized as: _ X _ X _ _ _ (where _ represents the remaining seats).
- We can use the "stars and bars" theorem to calculate the arrangements.

3. **Calculate the number of arrangements**:
- We can represent the arrangement as placing 2 persons in 4 available positions (after accounting for the gaps).
- The valid combinations can be calculated as follows:
- The number of ways to place 2 persons in 4 positions is given by the combination formula C(n, k) = n! / (k!(n-k)!)
- Here, n = 4 (the remaining seats) and k = 2 (the persons to place).
- Thus, the number of favorable outcomes = C(4, 2) = 6.

### Step 6: Calculate total outcomes
The total number of ways to choose 3 persons from 7 seats without any restrictions is given by:
C(7, 3) = 7! / (3!(7-3)!) = 35.

### Step 7: Calculate the probability
Now we can calculate the probability:
\[ P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{35}. \]

### Final Answer
The probability that the middle seat is always occupied and no two persons are consecutive is:
\[ \frac{6}{35}. \]