+0

# help plz

+1
97
10

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

Jan 13, 2019

#1
+533
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there are 10c2 or 45 games total

and there are 5c2 or 10 games where it is boy-v-boy

so it is 10/45 or 2/9

HOPE THIS HELPED!

Jan 13, 2019
#2
+18274
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We found from Chris' previous answer that girl vs girl  (or boy vs boy) is 10 games

Total games would be   10C2 = 45

10/45 = 2/9

Jan 13, 2019
#3
+101139
+2

As in your other question....the girls play 10 games against each other

And the boys play the same number

Each boy (or girl)  plays the opposite counterpart  5 * 5  = 25 times

So......the number of total games =  10 + 10 + 25  =  45

So....the fractional number of boy-boy games is   10 / 45   =  2 / 9

Jan 13, 2019
#4
+101139
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OK...by unanimous consensus.....2 / 9   it is.....!!!!!

Jan 13, 2019
edited by CPhill  Jan 13, 2019
#5
+533
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yep!!

Jan 13, 2019
#6
+101139
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LOL!!!!!

CPhill  Jan 13, 2019
#7
+533
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LOLOLOLOLOLOLOLOL

Jan 13, 2019
#8
+533
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i am going to post my own question; i can probably solve it on my own but i will post it anyway

Jan 13, 2019
#9
+101139
+1

OK......post away, asdf......!!!!!

CPhill  Jan 13, 2019
#10
+533
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i have posted the question cphill you can solve it if you want :P

Jan 13, 2019