Hi..I feel there might be something wrong with the problem.
Consider the 8 people as A,B,C,D,E,F,G,H
Now you can make following number of teams by choosing two people out of 8 people :
(8!)/ (6!) * (2!) = 28 teams.
I didn't know how to proceed after this so I tried to actually note down the teams :
AB,AC,AD,AE,AF,AG,AH,BC,BD,BE,BF,BG,BH,CD,CE,CF,CG,CH,DE,DF,DG,DH,EF,EG,EH,FG,FH,GH
or note it down this way :
AB AC AD AE AF AG AH
BC BD BE BF BG BH
CD CE CF CG CH
DE DF DG DH
EF EG EH
FG FH
GH
Now AB cannot play against team having A or B. So they can only play with teams having C,D,E,F,G or H. In short, AB can play against 15 teams. Similarly try for AC,AD,AE etc etc and you till see that all teams in first row can play with 15 teams and there are 7 teams in first row.
So total number of ways = 15 * 7 = 105.
Work downwards for each row and you get the following :
(15*7) + (10*6) + (6*5) + (4*3) + ( 3*1) = 210.
But this is not one of the options. So I probably made a mistake somewhere.
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