Six boys and six girls sit in a row randomly. The probability that the six girls sit together is

Options

A. `17/132`
B. `15/132`
C. `131/132`
D. `1/132`

Answer

To solve the problem of finding the probability that six girls sit together when six boys and six girls are seated randomly in a row, we can follow these steps:

### Step 1: Calculate the total arrangements of boys and girls.
The total number of ways to arrange 12 individuals (6 boys and 6 girls) in a row is given by the factorial of the total number of individuals.

\[ n(S) = 12! \]

### Step 2: Treat the six girls as a single unit.
If we want the six girls to sit together, we can treat them as one single unit or block. Therefore, we now have 7 units to arrange: 6 boys + 1 block of girls.

\[ n(A) = 7! \]

### Step 3: Arrange the girls within their block.
Within the block of girls, the 6 girls can be arranged among themselves in different ways. The number of arrangements of the 6 girls is given by:

\[ n(B) = 6! \]

### Step 4: Calculate the total arrangements where girls are together.
The total arrangements where the six girls sit together is the product of the arrangements of the blocks and the arrangements of the girls within their block:

\[ n(E) = n(A) \times n(B) = 7! \times 6! \]

### Step 5: Calculate the probability.
The probability that the six girls sit together is given by the ratio of the favorable arrangements to the total arrangements:

\[ P(E) = \frac{n(E)}{n(S)} = \frac{7! \times 6!}{12!} \]

### Step 6: Simplify the expression.
Now we can simplify the expression:

\[ P(E) = \frac{7! \times 6!}{12!} = \frac{7! \times 6!}{12 \times 11 \times 10 \times 9 \times 8 \times 7!} \]

The \(7!\) cancels out:

\[ P(E) = \frac{6!}{12 \times 11 \times 10 \times 9 \times 8} \]

Calculating \(6!\):

\[ 6! = 720 \]

Now substituting this value:

\[ P(E) = \frac{720}{12 \times 11 \times 10 \times 9 \times 8} \]

Calculating the denominator:

\[ 12 \times 11 = 132 \]
\[ 132 \times 10 = 1320 \]
\[ 1320 \times 9 = 11880 \]
\[ 11880 \times 8 = 95040 \]

So we have:

\[ P(E) = \frac{720}{95040} \]

### Step 7: Simplify the fraction.
Now we can simplify \( \frac{720}{95040} \):

\[ P(E) = \frac{1}{132} \]

### Final Answer:
Thus, the probability that the six girls sit together is:

\[ \boxed{\frac{1}{132}} \]