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.

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

Answer 1

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

Total of 45 games in the whole tournament.

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

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

Answer 2

Answer:

25

Step-by-step explanation:

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