in a tournament 8 teams have participated.in the 1st round each team will play with the other only once.top 4 will be qualified and after that knock out decision.how many total matches will be played and loser will play how many match??

• ek team dusri team ke saath sirf ek baar hi match khel skti h.it means 1st team can play 7 matches and 2nd team can play only 6 matches and so on.....so total matches played will be 28.now 4 qualified team will play on total 3 matches to decide the winner.so total matches played are 28+3=31.
  now the looser will play 7 matches + 2 matches from with that 4 qualified teams.so total wii be 9 matches.
• 13 years agoHelpfull: Yes(10) No(4)
• For 1st round no.of matches will be: 8C2 = (8*7)/2 = 28
  and for semifinal and final no of matches will be= 3
  so ans= 28+3=31
  
  a looser in final will play 7+2=9 Mathes
  
  
• 13 years agoHelpfull: Yes(10) No(0)
• for 1st round
  
  no of matches played by each team=7
  so total no of matches=7*8=56
  
  for knockout round
  
  4 teams r qualified
  so total no of matches=2
  
  for final
  1 match
  so total matches:56+2+1=59
  
  now for loser
  no of matches played in 1st round=7
  so loser will play 7 matches
• 13 years agoHelpfull: Yes(9) No(11)
• For Round Round 8Th Team Will Play With Remaining 7 Teams...Nxt 7Th Team Will Play Only With 6 Teams As It ALready Played With 8Th Team N One Team Plays With ANoter Only Once...In Thz Manner Total Matches Plyed Ra 7+6+5+4+3+2+1=28 iN fIRST rOUND...In Semifinals 2 Matches Final 1 Match Total 28+2+1=31 Matches
• 13 years agoHelpfull: Yes(4) No(2)
• since it is knockout match then only 1 chance for each team so
  2^8=256
• 13 years agoHelpfull: Yes(0) No(3)