If there are 6 girls and 5 boys who sit in a row, then the probability that no two boys sit together is

Options

A. `(6! 6!)/( 2 ! 11 !)`
B. `(7 ! 5!)/(2 ! 11!)`
C. `(6 ! 7!)/(2 ! 11!)`
D. None of these

Answer

To solve the problem of finding the probability that no two boys sit together when 6 girls and 5 boys are seated in a row, we can follow these steps:

### Step 1: Calculate the total number of arrangements of 11 people.
The total number of people is 6 girls + 5 boys = 11 people. The total arrangements of these 11 people is given by:
\[ 11! \]

### Step 2: Arrange the girls.
First, we arrange the 6 girls. The number of ways to arrange 6 girls is:
\[ 6! \]

### Step 3: Identify the available positions for boys.
When the 6 girls are arranged, they create gaps where the boys can sit. The arrangement of 6 girls creates 7 possible gaps (one before each girl and one after the last girl). The gaps can be visualized as follows:
- _ G1 _ G2 _ G3 _ G4 _ G5 _ G6 _

So, there are 7 positions available for the boys.

### Step 4: Choose positions for the boys.
We need to select 5 positions out of these 7 for the boys. The number of ways to choose 5 positions from 7 is given by:
\[ \binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 5: Arrange the boys in the chosen positions.
Once we have chosen the 5 positions for the boys, we can arrange the 5 boys in these positions. The number of ways to arrange 5 boys is:
\[ 5! \]

### Step 6: Calculate the total favorable arrangements.
The total number of favorable arrangements where no two boys sit together is the product of the arrangements of girls, the ways to choose positions for boys, and the arrangements of boys:
\[ \text{Favorable arrangements} = 6! \times \binom{7}{5} \times 5! \]

### Step 7: Calculate the probability.
The probability that no two boys sit together is given by the ratio of the number of favorable arrangements to the total arrangements:
\[ P = \frac{6! \times \binom{7}{5} \times 5!}{11!} \]

### Step 8: Simplify the probability.
Substituting the values we calculated:
\[ P = \frac{6! \times 21 \times 5!}{11!} \]
We can express \(11!\) as \(11 \times 10 \times 9 \times 8 \times 7 \times 6!\):
\[ P = \frac{6! \times 21 \times 5!}{11 \times 10 \times 9 \times 8 \times 7 \times 6!} \]
Cancelling \(6!\) gives:
\[ P = \frac{21 \times 5!}{11 \times 10 \times 9 \times 8 \times 7} \]

### Final Result:
Thus, the final probability that no two boys sit together is:
\[ P = \frac{7! \times 6!}{2 \times 11!} \]