### Akash_Roy's blog

By Akash_Roy, history, 5 weeks ago,

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 ?

Sorry for my poor number formatting.

• +3

 » 5 weeks ago, # |   0 If someone knows how to enter numbers and symbols(like in editorial), kindly share in the comments section. It will help in future :).
•  » » 5 weeks ago, # ^ |   0 Its LaTeX
•  » » » 5 weeks ago, # ^ |   0 Thanks slxsh . It helps .
 » 5 weeks ago, # |   0 (n-1)! gives the total number of ways you can possibly arrange n people in a circular permutation. Why divide by 2 and not 4? Dividing by 2, is for not counting the same arrangement for both dance 1 and dance 2. suppose (2,1) and (3,4) is a way to arrange them. But, (3,4) and (2,1) is also a way that we are counting. but we consider these the same arrangement. And hence we divide by 2.
•  » » 5 weeks ago, # ^ |   0 Thanks nagaraj41 . I got your point now :)