The 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 [2023]

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Solution

Q.
The 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 [2023]

1 $7 {(360)}^{2}$

2 $126 {(5!)}^{2}$

3 $7 {(720)}^{2}$

4 $720$

Ans.
(2)

Number of girls = 5

Number of boys = 7

The number of ways of arranging boys around a table is (7 – 1)! = 6!

Now, there are 7 spaces in between two boys, so 5 girls arranged in 7 gaps by ${}^{7}{P_{5}}$ ways.

So, required number of ways = $6! \times {}^{7}{P_{5}} = 126 \times {(5!)}^{2}$

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