The number of ways of arranging 9 men and 5 women around a circular table so that no two women come together…
The number of ways of arranging 9 men and 5 women around a circular table so that no two women come together are
$8!^8 P_5$
$9!{ }^9 P_5$
$8!{ }^9 P_5$
$8!5$ !
Solution
First fix the men in circular arrangement, which can be done in $(9-1)!=8$ ! ways
Now, there are 9 places between the men and 5 womens to be seated between the men. That can be done in ${ }^9 \mathrm{P}_5$ ways.
$\therefore$ Total no. of ways to sit $=8!{ }^9 \mathrm{P}_5$