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.
+30 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.
+37153 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
