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Miscellaneous Exercise · Q9

Q.Number of ways in which 15 different children can sit in a merry-go-round relative to one another is

(i) 12(14!)\dfrac{1}{2}(14!)
(ii) 14!14!
(iii) 12(15!)\dfrac{1}{2}(15!)
(iv) 2×14!2 \times 14!
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Seating 15 distinct children "relative to one another" on a merry-go-round is a circular arrangement where clockwise and anticlockwise orders are treated as the same, halving the usual circular-permutation count.

Circular permutations of nn distinct objects =(n−1)!=(n-1)! (one position is fixed to remove the ring's rotational symmetry). When the arrangement viewed from the front is indistinguishable from the one viewed from the back (clockwise ≡\equiv anticlockwise), further divide by 2: (n−1)!2\dfrac{(n-1)!}{2}.

  1. There are n=15n=15 distinct children. Fix any one child's seat to eliminate the rotational symmetry of the circular arrangement; the remaining n−1=14n-1=14 children can then be seated in the other 14 positions in 14!14! ways. …

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