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It is required to divide the 2n members of a club into n disjoint teams of 2 members each. The teams are not labelled. The number of ways in which this can be done is:


(A) (2n)! / 2n 


 (B) (2n)! / n! 


(C) (2n)! / 2n · n!


(D) n! / 2

closed as a duplicate of: TIFR2012-A-7
asked in Combinatory by Active (4.9k points)
closed by | 37 views

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