Five boys and four girls sit in a row randomly. The probability that no two girls sit together

Options

A. `41/42`
B. `5/42`
C. `11/42`
D. `3/21`

Answer

To solve the problem of finding the probability that no two girls sit together when five boys and four girls sit in a row randomly, we can follow these steps:

### Step 1: Calculate Total Arrangements
First, we need to find the total number of arrangements of 5 boys and 4 girls. The total number of people is 9 (5 boys + 4 girls).

The total possible arrangements of 9 people (5 boys and 4 girls) can be calculated using the factorial of the total number of people:
\[ \text{Total arrangements} = 9! \]

### Step 2: Arrange Boys
Next, we arrange the boys first. The boys can be arranged among themselves in:
\[ \text{Arrangements of boys} = 5! \]

### Step 3: Identify Positions for Girls
When the boys are arranged, they create gaps where the girls can sit. For 5 boys, there are 6 potential gaps (one before each boy, one after the last boy):
- _ B _ B _ B _ B _ B _

This means there are 6 gaps available for placing the girls.

### Step 4: Choose Positions for Girls
To ensure that no two girls sit together, we need to select 4 out of these 6 gaps to place the girls. The number of ways to choose 4 gaps from 6 is given by the combination formula:
\[ \text{Ways to choose gaps} = \binom{6}{4} \]

### Step 5: Arrange Girls in Chosen Gaps
Once we have chosen the gaps, we can arrange the 4 girls in those selected gaps. The girls can be arranged among themselves in:
\[ \text{Arrangements of girls} = 4! \]

### Step 6: Calculate Favorable Outcomes
Now, we can calculate the total number of favorable outcomes where no two girls sit together:
\[ \text{Favorable outcomes} = \text{Arrangements of boys} \times \text{Ways to choose gaps} \times \text{Arrangements of girls} \]
\[ = 5! \times \binom{6}{4} \times 4! \]

### Step 7: Calculate Probability
Finally, the probability that no two girls sit together is given by the ratio of the number of favorable outcomes to the total arrangements:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total arrangements}} \]
\[ = \frac{5! \times \binom{6}{4} \times 4!}{9!} \]

### Step 8: Simplify the Expression
Now we can simplify this expression:
1. Calculate \(5! = 120\)
2. Calculate \(4! = 24\)
3. Calculate \(\binom{6}{4} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15\)
4. Thus, \(Favorable outcomes = 120 \times 15 \times 24\)
5. Calculate \(9! = 362880\)

Putting it all together:
\[ \text{Favorable outcomes} = 120 \times 15 \times 24 = 43200 \]
\[ \text{Probability} = \frac{43200}{362880} = \frac{5}{42} \]

### Final Answer
The probability that no two girls sit together is:
\[ \frac{5}{42} \]