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

Question

The number of ways in which 5 ladies and 7 gentleman can be seated in round table so that no two ladies together, is (A) (7/2)(720)2 (B) 7(360)2 (C) 7(720)2 (D) 720

Tardigrade Admin · Accepted Answer

First we fix the alternate position of 7 gentlemen in a round table by 6! ways. There are seven positions between the gentlemen in which 5 ladies can be seated in 7p5 ways <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2ae5e0384f59a8372c8e91589fd3371d-.png" /> ∴ required number of ways =6!×(7!/2) =(7/2)(720)2

Q. The number of ways in which $5$ ladies and $7$ gentleman can be seated in round table so that no two ladies together, is

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

$7 (360)^{2}$

$7 (720)^{2}$

$720$

First we fix the alternate position of $7$ 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 $= 6! \times \frac{7 !}{2}$
$= \frac{7}{2} (720)^{2}$

Q. The number of ways in which 5 ladies and 7 gentleman can be seated in round table so that no two ladies together, is

27​(720)2

7(360)2

7(720)2

720