In how many ways can you arrange $4$ men and $4$ women in a row of $8$ seats if one man and a woman will insist not to be seated together?
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HINT : Arrange 3 men and 3 women (assuming these six people don't mind sitting anywhere) in $6!$ ways and in the 7 places got after arranging six people, choose 2 places and place the 2 people who insist not being together in $7 \choose 2$ * $2! $ ways
answer would be $6!$ $\times$ $7 \choose 2 $ $\times 2! $ = $30240$
We have $4$ men and $4$ women, and we cannot have $2$ members of the same gender sitting next to each other. So, for the first seat, we have $8$ possible choices. The second seat has $4$ possible choices (the $4$ corresponding to the opposite gender from the individual chosen in the first seat); the third and fourth seats both have $3$ possible choices; the fifth and sixth seats both have $2$ possible choices; and one choice exists for both the seventh and eighth seat. Hence, we have $184.108.40.206.220.127.116.11=1152$ different ways to alternate the males and females.