Question

The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together, is

• A. $8!$
• B. $4!$
• C. $8!4!$
• D. $7{!}^8{P}_4$

Answer 1

The correct option is D $7{!}^8{P}_4$

Numbers of ways such that no two women. sit together is such that to first arrange all the men in a circle and then place the women in places between them

number of ways of arranging 8 men in a circle is $(n-1)!=(8-1)!=7!$

number of ways of placing 4 women in 8 places between men is ${}^8{P}_4$

$\therefore$ Total numbers of ways${=}^8{P}_4\times 7!$