Question

In how many ways $7$ men and $7$ women can be seated around a round table such that no two women can sit together?

• A. $(7!{)}^2$
• B. $7!\times 6!$
• C. $(6!{)}^2$
• D. $7!$

Answer 1

Answer: (B)

$7!\times 6!$

Lets first place the men (M). * here indicates the linker of round table *M -M - M - M - M* which is in $(7-1)!$ ways = $6!$
So we have to place the women in between the men which is on the $5$ empty seats So $7$ women can sit on $7$ seats in $\left(7\right)!$ ways
So the answer is $7!\times 6!$