In sub - part (i) to (x) choose the correct option and in sub - part (xi) to (xy), answer the questions as intructed .
Six boys and six girls sit in row rouldelary Probability that boys and girl sit alternately is :

Options

A. `(1)/(462)`
B. `(1)/(264)`
C. `(1)/(426)`
D. None of these

Answer

To solve the problem of finding the probability that boys and girls sit alternately in a circular arrangement, we can follow these steps:

### Step 1: Understand the arrangement
We have 6 boys and 6 girls, making a total of 12 individuals. Since we want them to sit alternately, we can either start with a boy or a girl.

### Step 2: Calculate the total arrangements
In a circular arrangement, the total number of ways to arrange 12 individuals is given by \((n-1)!\) where \(n\) is the total number of individuals. Thus, the total arrangements are:
\[ \text{Total arrangements} = (12 - 1)! = 11! \]

### Step 3: Calculate the favorable arrangements
For the boys and girls to sit alternately, we can consider two cases:
1. Starting with a boy: The arrangement will be BGBGBGBGBGBG.
2. Starting with a girl: The arrangement will be GBGBGBGBGBG.

In both cases, we can arrange the boys and girls independently. The number of ways to arrange 6 boys is \(6!\) and the number of ways to arrange 6 girls is also \(6!\).

Thus, the favorable arrangements for each case are:
\[ \text{Favorable arrangements for one case} = 6! \times 6! \]
Since there are two cases (starting with a boy or starting with a girl), the total favorable arrangements are:
\[ \text{Total favorable arrangements} = 2 \times (6! \times 6!) \]

### Step 4: Calculate the probability
The probability \(P\) that boys and girls sit alternately is given by the ratio of favorable arrangements to total arrangements:
\[ P = \frac{\text{Total favorable arrangements}}{\text{Total arrangements}} = \frac{2 \times (6! \times 6!)}{11!} \]

### Step 5: Simplify the probability
We know that \(11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6!\). Therefore, we can simplify:
\[ P = \frac{2 \times (6! \times 6!)}{11 \times 10 \times 9 \times 8 \times 7 \times 6!} = \frac{2 \times 6!}{11 \times 10 \times 9 \times 8 \times 7} \]

### Step 6: Calculate \(6!\)
Calculating \(6!\):
\[ 6! = 720 \]

### Step 7: Substitute and calculate the final probability
Substituting \(6!\) into the probability:
\[ P = \frac{2 \times 720}{11 \times 10 \times 9 \times 8 \times 7} \]
Calculating the denominator:
\[ 11 \times 10 = 110, \quad 110 \times 9 = 990, \quad 990 \times 8 = 7920, \quad 7920 \times 7 = 55440 \]
Thus, the probability becomes:
\[ P = \frac{1440}{55440} = \frac{1}{38.5} \approx \frac{1}{462} \]

### Final Answer
The probability that boys and girls sit alternately is:
\[ \frac{1}{462} \]