In a tournament of table tennis, a team consists of 2 member

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.

@everything's eventual: I think your answer is correct and bravo!! for the effort.

As we have to distribute 8 persons in 2 groups, No. of ways of doing that will be = 8C2.6C2 ways = 420 ways
However, any given arrangement will repeat itself for both the teams
e.g., Team 1 Team 2
(1,2) (3,4)
(3,4) (1,2)
To get rid of repetition, divide 420 by 2, which gives us 210 ways, which is our answer.

That was awesome !! Both of you , highly appreciated , yes the OA was given as (C) 630, which caused me to put this question up for discussion , I think 210 is the right answer.

Thank you both of you.