The eight members of the chess club have a tournament in which every player plays a game against every other player. How many games are there in the tournament?

Question

The eight members of the chess club have a tournament in which every player plays a game against every other player. How many games are there in the tournament?

Answer 1

Hi Sharon,

 

This answer is wrong. Sorry. But I decided to not edit it out because it serves as an example of a common mistake. I double counted so much.

 

Every player (1) plays with every other player (7 players in all that he/she plays with) and you have 8 of those. => 8*7 = 56 games 

Answer 2

This is a combination problem (not permutation, because order doesn't matter...A playing B is the same as B playing A).

 

So you just want ₈C₂

 

• ₈C₂= 8!/[(2!)(8-2)!]
• (8*7*6*5*4*3*2*1)/[(2*1)(6*5*4*3*2*1]
• (8*7)/2    ~  Note everything that easily cancels here!
• 56/2
• 28 matches

 

You can also do this with a decision tree thing and a bit of logic. 

 

• 1st player plays all 7 other players, for 7 matches
• 2nd player only needs 6 more matches, as his game with the 1st player is already accounted for above.
• 3rd player...5 matches (similar logic)
• 4th player...4 more matches
• 5th...3
• 6th...2
• 7th...1
• 8th, no new matches, as he's already accounted for with the matches above.
• TOTAL MATCHES = 7+6+5+4+3+2+1=28 matches

Answer 3

there are 8 members

call them A  B  C  D  E  F  G  H

A plays B,C,D,E,F,G, and H for a total of 7 games

B plays C,D,E,F,G, and H for a total of 6 games (A and B already played !!)

C plays D,E,F,G, and H for a total of 5 games (C already played A and B !!)

D plays E,F,G, and H for a total of 4 games

E plays F,G, and H for a total of 3 games

F plays G and H for a total of 2 games

G plays H for a total of 1 game

H, everybody played H already !!!

add 1,2,3,4,5,6, and 7

1+2+3+4+5+6+7=28 games in the tournament

I hope you saw the pattern before I finished the problem !!!