Question

$6$ men and $6$ women are to sit round a table such that no two woman are sitting together. The number of seating arrangements is:

• A. $(5!{)}^2$
• B. $(6!{)}^2$
• C. $5!.6!$
• D. $6!$

Answer 1

The correct option is C $5!.6!$

Number of possible ways in which women can seat= $5!$
now $6$ places is created so Number of possible ways in which man can seat=$6!$
So, total number of seating arrangement= $5!\times 6!$