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

Explanation

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 arrange the 5 boys. The number of ways to arrange 5 boys in a row is given by the factorial of the number of boys:

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

### Step 2: Identify Positions for Girls
Once the boys are arranged, we need to find the positions available for the girls. When the 5 boys are seated, they create 6 potential positions for the girls (one before each boy, one after the last boy):

- Position 1: Before Boy 1
- Position 2: Between Boy 1 and Boy 2
- Position 3: Between Boy 2 and Boy 3
- Position 4: Between Boy 3 and Boy 4
- Position 5: Between Boy 4 and Boy 5
- Position 6: After Boy 5

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

\[ \text{Ways to choose positions for girls} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

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

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

### Step 5: Calculate Total Arrangements
Finally, we multiply the number of ways to arrange the boys, the number of ways to choose positions for the girls, and the number of ways to arrange the girls:

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

Calculating this gives:

\[ 120 \times 20 = 2400 \]
\[ 2400 \times 6 = 14400 \]

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 **14,400**.