Question 351011
there are 2n teams and n games, each game have 2 team...

for first match team can be assigned by  2nC2 ways
    
 now there are only (2n -2) team 

     for second game team can be assigned by (2n-2)C2 ways  

     for third game ....................... (2n-4)C2 ways 
     
       for fourth game .................... (2n-6)C2 ways
                                          
                                                  and so on.


total no. of ways = 2nC2 * (2n-2)C2 * ..............* 4C2 * 2C2 
                   
                  
after solving it we will find  (2n)! / (2*2*2...n times)
                                  
                              = (2n)! / (2^n)


Sometimes there may be typing mistake in solution of a problem, so please ignore it.
Understand the concept and try the problem yourself.