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)

$7! \times 6!$

Answer 2

Fix up 1 man and the remaining 6 men can be seated in 6! ways. Now no two women are to sit together and as such the 7 women are to be arranged in seven empty seats between two consecutive men and number of arrangement will be 7!. Hence by fundamental theorem the total number of ways

= 7! × 6!.