+0  
 
0
42
4
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Part A)   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. Of all the matches, how many of them have a left-handed player competing against a right-handed player?

 

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

 

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

 May 24, 2021
 #1
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A) There are 50 matches.

 

B) There are 50 matches.

 

C) The fraction is 1/3.

 May 24, 2021
 #2
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part a and part b are wrong :(

Guest May 24, 2021
 #3
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10 people each have to play 9 other people = 90     BUT TWO people participate in each match   

                      so  90/2 = 45 matches

 May 24, 2021
 #4
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Part A is 25, part B is 10!

 May 25, 2021

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