0 votes 0 votes In how many ways $8$ girls and $8$ boys can sit around a circular table so that no two boys sit together? $(7!)^2$ $(8!)^2$ $7!8!$ $15!$ Combinatory nielit2017dec-scientistb discrete-mathematics circular-permutation + – admin asked Mar 30, 2020 retagged Oct 23, 2020 by Krithiga2101 admin 2.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Anu007 commented Dec 18, 2017 i edited by stblue Dec 18, 2017 reply Follow Share number of ways to sit boys first = (n-1)! = 7! now girls can sit in = 8! Total = 7! * 8! 2 votes 2 votes Srken commented Oct 30, 2022 reply Follow Share This problem can be considered as boys and girls sitting alternatively also.So why don't we apply 7!7! . According to circular arrangement if only the positions are changed then all of the arrangements are considered as same right , then Why do we only take boys as 7! And why can't we add girls also as 7! . 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes First arrange boys or girls on the circle No of ways to arrange in nplaces around circle= (n-1)! Here let say first we arrange boys in 8 places= (8-1)=7! Now in the remaing 8 places we have to arrange girls= 8! So total ways = 7! * 8!. Answer:C Dileep kumar M 6 answered Dec 18, 2017 Dileep kumar M 6 comment Share Follow See all 0 reply Please log in or register to add a comment.