The number of ways in which 5 ladies and 7 gentlemen can be seated in a round table so that no two ladies sit together, is

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The number of ways in which 5 ladies and 7 gentlemen can be seated in a round table so that no two ladies sit together, is

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$\frac{7}{2} (720)^{2}$

7(360)²

7(720)²

720

answer is A.

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Detailed Solution

First we fix the alternate positions of z gentlemen in a round table by 6! ways.

There are seven positions between the gentlemen in which 5 ladies can be seated in $^{7} P_{5}$ . ways.
Required number of ways

$= 6! \times \frac{7!}{2!} = \frac{7}{2} (720)^{2}$

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