+0  
 
+1
110
6
avatar+434 

There are 5 girls and 5 boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once.

What fraction of the games are boy-versus-boy? Enter your answer as a fraction in simplified form.

SmartMathMan  Jan 17, 2018
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6+0 Answers

 #1
avatar+85894 
0

Boy-Boy  =  C(5,2)  =  10 games

Girl-Girl = C(5,2)  =  10 games

Boy-Girl  =  5 * 5  =  25 games

 

So..... Boy - Boy  =

 

[ 10 ]  /  [ 10 + 10 + 25 ]  =

 

10 / 45  =

 

2 / 9

 

 

 

cool cool cool

CPhill  Jan 17, 2018
edited by CPhill  Jan 17, 2018
edited by CPhill  Jan 17, 2018
 #2
avatar+434 
+1

that was incorrect but i dont know why

SmartMathMan  Jan 17, 2018
edited by SmartMathMan  Jan 17, 2018
 #3
avatar+434 
+1

i figured it out you simplified it wrong

SmartMathMan  Jan 17, 2018
 #4
avatar+434 
+1

as a side note both problems i needed help with you got wrong but only because of little mistakes. Maybe pay a little more attetion? thx for your help i just needed some and i figured it out anyway. I hope that feedback helped

SmartMathMan  Jan 17, 2018
 #5
avatar+85894 
0

Fixed now....I'm trying to answer too many questions at once....LOL!!!

 

 

cool cool cool

CPhill  Jan 17, 2018
 #6
avatar+434 
+1

You fixed it. Thx i know. I dont want to  be complaining rather helping. its a struggle i can tell. thx for all the help! :-)

SmartMathMan  Jan 17, 2018
edited by SmartMathMan  Jan 17, 2018
edited by SmartMathMan  Jan 17, 2018

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