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. $\frac{7}{2}(720)^2$
• B. $7(360)^2$
• C. $7(720)^2$
• D. $720$

Answer 1

Answer: (A)

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

$\therefore $ required number of ways $=6!\times\frac{7!}{2}$
$=\frac{7}{2}\left(720\right)^{2}$