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We select members for first round dance. Then members of second round dance get automatically selected . Hence the factor nC(n/2). Then we arrange both the round dance. This leads to nC(n/2)*(n-1)!*(n-1)! {Circular permutation}. After this why are we dividing the answer only by 2? I feel we should divide final answer by 4 since there are two round dances. Since clockwise and anticlockwise arrangement does not matter, the final answer should be [nC(n/2)*(n-1)!*(n-1)!]/4. Why is it [nC(n/2)*(n-1)!*(n-1)!]/2 ?

Please help.

Sorry for my poor number formatting.