In how many ways can 5 boys and 3 girls sit in a row so that no two girls are together ?

Options

A. `5!xx3!`
B. `.^(4)P_(3)xx5!`
C. `.^(6)P_(3)xx5!`
D. `.^(5)P_(3)xx3!`

Answer

To solve the problem of how many ways 5 boys and 3 girls can sit in a row such that no two girls are together, we can follow these steps:

### Step 1: Arrange the Boys
First, we need to arrange the 5 boys. The number of ways to arrange 5 boys is given by the factorial of the number of boys.

\[ \text{Ways to arrange boys} = 5! = 120 \]

### Step 2: Identify Positions for the Girls
Once the boys are arranged, we can identify the positions where the girls can sit. When 5 boys are arranged, they create 6 potential gaps for the girls to sit: one before the first boy, one between each pair of boys, and one after the last boy.

The arrangement looks like this:
- _ B1 _ B2 _ B3 _ B4 _ B5 _

This gives us 6 gaps (denoted by underscores) where the girls can be placed.

### Step 3: Choose Positions for the Girls
We need to choose 3 out of these 6 gaps for the girls to sit in. The number of ways to choose 3 gaps from 6 is given by the combination formula:

\[ \text{Ways to choose gaps} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = 20 \]

### Step 4: Arrange the Girls
After choosing the gaps, we can arrange the 3 girls in those chosen gaps. The number of ways to arrange 3 girls is given by the factorial of the number of girls.

\[ \text{Ways to arrange girls} = 3! = 6 \]

### Step 5: Calculate Total Arrangements
Now, we can calculate the total number of arrangements by multiplying the number of ways to arrange the boys, the number of ways to choose the gaps, and the number of ways to arrange the girls.

\[ \text{Total arrangements} = (\text{Ways to arrange boys}) \times (\text{Ways to choose gaps}) \times (\text{Ways to arrange girls}) \]

Substituting the values we calculated:

\[ \text{Total arrangements} = 5! \times \binom{6}{3} \times 3! = 120 \times 20 \times 6 \]

Calculating this gives:

\[ \text{Total arrangements} = 120 \times 20 = 2400 \]
\[ 2400 \times 6 = 14400 \]

Thus, the total number of ways in which 5 boys and 3 girls can sit in a row such that no two girls are together is **14400**.