Question

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

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

Answer 1

Answer: (A)

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

Answer 2

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