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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 Loyal (3.7k points)
closed by | 17 views


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