The number of way ,in which `5` girls and `7` boys can be seated at a round table so that no two girls sit together , is

Options

A. `126(5!)^(2)`
B. `7(360)^(2)`
C. `7(720)^(2)`
D. `720`

Answer

To solve the problem of seating 5 girls and 7 boys at a round table such that no two girls sit together, we can follow these steps:

### Step-by-Step Solution:

1. **Arrange the Boys:**
Since we are seating them at a round table, we can fix one boy to eliminate the effect of rotations. This means we can arrange the remaining 6 boys in a linear fashion. The number of ways to arrange 7 boys in a circle is given by (7-1)! = 6!.

\[ \text{Ways to arrange boys} = 6! = 720 \]

2. **Create Gaps for Girls:**
When the 7 boys are seated, they create 7 gaps (one between each pair of boys and one on each end) where the girls can be seated. We need to ensure that no two girls sit together, which means we can only place one girl in each gap.

3. **Choose Gaps for Girls:**
We need to select 5 out of these 7 gaps to place the girls. The number of ways to choose 5 gaps from 7 is given by the combination formula \( \binom{n}{r} \).

\[ \text{Ways to choose gaps} = \binom{7}{5} = \binom{7}{2} = 21 \]

4. **Arrange the Girls:**
Once the gaps are chosen, we can arrange the 5 girls in those gaps. The number of ways to arrange 5 girls is given by 5!.

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

5. **Total Arrangements:**
Finally, to find the total number of arrangements, we multiply 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.

\[ \text{Total arrangements} = 6! \times \binom{7}{5} \times 5! = 720 \times 21 \times 120 \]

Now we calculate this:

\[ 720 \times 21 = 15120 \]
\[ 15120 \times 120 = 1814400 \]

Thus, the total number of ways in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together is **1,814,400**.