Question

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.

Answer 1

The total number of games a girl play against a boy is 25, the number of games a girl plays against a boy is 10, the fraction of the games are boy-versus-boy is 2/9, and 1/5 is the fraction of all games at the tournament Anita plays.

How to carry out Permutation and Combination?

We are given that there are 5 girls and 5 boys in a chess club.

Thus, total games that girls play against a boy = 5×5 = 25

The number of games a girl play against another is the total number of games = 10

The number of games that are boy vs boy = 10

The number of games Boy vs Boy + Girl vs Girl + Boy vs Girl = 10 + 10 + 25 =  45

Fraction of the games that are boy vs boy = 10/45 = 2/9

If Anita is one of the club members, then Anita will have to play against 9 people at the tournament.

Total number of games = 45

Then the fraction of all games at the tournament Anita play the game.= 9/45 ⇒ 1/5

Thus, the total number of games a girl play against a boy is 25, the number of games a girl plays against a boy is 10, the fraction of the games are boy-versus-boy is 2/9, and 1/5 fraction of all games at the tournament Anita plays.

Read more about Permutation and Combination at; brainly.com/question/4658834

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