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.

To get the total number of games possible, we will use the combination formula.
Total number of games : [tex] _{10}C_{2} [/tex] = 45

Total of 45 games in the whole tournament.

Meanwhile, if there would be boy-versus-boy, we will have [tex]_{5}C_{2} [/tex]
The answer is 10.

The fraction for boy-versus-boy game will be 10/45 or 2/9 when simplified.

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Kesha3723 Kesha3723 · Answer 2 · 2021-05-08T17:09:48+02:00

Kesha3723 Kesha3723

08-05-2021

Answer:

25

Step-by-step explanation:

The answer above me is wrong, for which I have actually did the question, the answer is 25

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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.

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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.