Question

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.
In how many games does a girl play against a boy?(answer:25)

In how many games does a girl play against another girl?(answer:10)

What fraction of the games are boy-versus-boy? Enter your answer as a fraction in simplified form.(answer:2/9)

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.(answer: I have no idea)
(this single question worth 64 points)

Answer 1

1. In how many games does a girl play against a boy?
A: Because there are 5 girls, each will play against 5 boys, so in total is 5*5 = 25 times.

2. In how many games does a girl play against another girl?
A: In this case, we only need to calculate how many girl pairs are there among these 5 girls, and it become a combination problem which can solved by $C^{2}_{5}=\frac{5*4}{2*1}=10$

3.The fraction of games "B vs B" should be the number of "B vs B" (which is 10) divide the total number of the games "B vs B" & "G vs G" & "B vs G" (which is 10+10+25=45, see the analysis in 1.A and 2.A), so the fraction should be 10/45=2/9

4.One person will have to play against 9 people at the tournament, and the total number of the games is 45, so the fraction should be 9/45 = 1/5

Answer 2

Answer:

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