There are 2 vans each having numbered seats, 3 in the front and 4 at the back. There are 3 girls and 9 boys to be seated in the vans. The probablity of 3 girls sitting together in a back row on adjacent seats, is

Options

A. `(1)/(13)`
B. `(1)/(39)`
C. `(1)/(65)`
D. `(1)/(91)`

Answer

To solve the problem of finding the probability that 3 girls sit together in the back row of the vans, we can follow these steps:

### Step 1: Understand the seating arrangement
Each van has:
- 3 seats in the front
- 4 seats in the back

Thus, there are a total of 14 seats (2 vans × (3 front + 4 back) = 14 seats).

### Step 2: Total number of students
We have:
- 3 girls
- 9 boys

This gives us a total of 12 students (3 girls + 9 boys).

### Step 3: Calculate total arrangements of students
The total number of ways to arrange 12 students in 14 seats is given by the permutation formula:

\[ \text{Total arrangements} = P(14, 12) = \frac{14!}{(14-12)!} = \frac{14!}{2!} = \frac{14!}{2} = 14 \times 13 \times 12! \]

### Step 4: Choose back seats for the girls
Since we want the girls to sit together in the back row, we first choose one of the two vans (2 choices).

### Step 5: Arranging the girls
The 3 girls can sit together in the back row. We can treat the 3 girls as a single unit or block. This block can occupy 3 adjacent seats in the back row. The arrangement of the girls within this block can be done in \(3!\) ways.

### Step 6: Arranging the boys
After seating the girls, we have 9 boys left to seat. The remaining seats will be filled by the boys. Since one van has 4 seats in the back and we have already seated the girls, we have 1 remaining seat in that van and 3 seats in the other van. Thus, we have 4 seats left for the boys.

The number of ways to arrange the boys in the remaining seats is given by:

\[ P(11, 9) = \frac{11!}{(11-9)!} = \frac{11!}{2!} = \frac{11!}{2} \]

### Step 7: Calculate the favorable arrangements
The total number of favorable arrangements where the 3 girls sit together in the back row is:

\[ \text{Favorable arrangements} = 2 \times 3! \times P(11, 9) = 2 \times 6 \times \frac{11!}{2} = 6 \times 11! \]

### Step 8: Calculate the probability
The probability \(P\) that the 3 girls sit together in the back row is given by the ratio of favorable arrangements to total arrangements:

\[ P = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{6 \times 11!}{\frac{14!}{2}} = \frac{6 \times 11! \times 2}{14!} \]

### Step 9: Simplifying the probability
We can simplify this further:

\[ P = \frac{12 \times 6}{14 \times 13 \times 12} = \frac{72}{1820} = \frac{1}{91} \]

Thus, the probability of the 3 girls sitting together in the back row on adjacent seats is:

\[ \boxed{\frac{1}{91}} \]