Of the 10 kids in a chess club, 5 are left-handed and 5 are right-handed. The club holds a round-robin tournament in which every player plays against every other player exactly once.

1. Of all the matches, how many of them have a left-handed player competing against a right-handed player?


2. Of all the matches, how many of them have two left-handed players?


3. What fraction of the games have two right-handed players? Enter your answer as a fraction in simplified form.


4. If Anita is one of the club members, what fraction of all games at the tournament does Anita play in? Enter your answer as a fraction in simplified form.

 Jun 13, 2021
edited by Guest  Jun 13, 2021

We know there are 55 total games in the tournament so keep that in mind.


1. There are 5 left-handed players and 5 right-handed players so each left-handed player will play 5 right-handed players so there are 5*5 matches in there so it is 25.


2. Well, that would just be 4+3+2+1 left-hand only matches which will add up to 10 rounds.


3.  It would be the same as the left-handed players which is 10/55 which simplifies to 2/11.


4. Anita will play 9 games so it is 9/55 which cannot be simplified. 

 Jun 13, 2021

Not sure if Ninchux is correct....or if my answer is correct either...but here it is


10 players must each play 9 others  = 90 games BUT two players are in each game   90 / 2 = 45 games total


5 players L  must each play the other 5 players R =   25 games with   L vs R


5 L must each play the other 4 lefties   = 20  but two players in each game means  10 games with L vs L


Two right handed players would also be 10 games out of total 45        10/45 = 2/9


Anita must play 9 other players in the tournament    9 /45 =   1/5 of the games

 Jun 13, 2021

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