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Question- $5$ boys and $5$ girls to be arranged in a row such that boys and girls are alternate and a particular boy and a particular girl are never together.

Try- taking the total number of cases to be $5!×5!×2$ then making groups such as $b2g2$ $b3g3$ $b4g4$ $b5g5$ and subtracting the case of permutations of b1g1 from original .

Is this the right and an answer would be very much appreciated.

Jessica Tiberio
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user584728
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1 Answers1

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Your $2 \cdot 5!^2$ is correct for the number of ways to seat them ignoring the restriction of the two who won't sit together. Now subtract the number of ways with them sitting together. The first one you come to in the row can be either the boy or girl and can be in nine places.

Ross Millikan
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  • Yup thought so but this does not lead to the answer ans as u r ignoring the possibility of arrangement of the group b1g1 in that way it may not be alternate in all permutations – user584728 May 09 '18 at 15:15
  • You need to compute the number of alternating permutations where the couple sits together. Pick where they sit first, then seat the rest of the people in the other chairs, making sure they alternate. – Ross Millikan May 09 '18 at 16:41
  • @user584728 Choosing the position where the first member of the couple sits and whether the boy or girl comes first determines the relative positions of the other boys and girls. If the couple sits in the sixth and seventh seats and the boy sits in the sixth seat, then the row begins with a girl. – N. F. Taussig May 10 '18 at 11:03