There are girls and 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.

Note: We need the number of girls and boys to calculate this question.
This is the application of combinations;
Suppose we have 6 boys and 6 girls participating in the tournament ;
take C to represent the combination:
The total number of games = 12 C 2= 66
only boys versus boys games= 6 C 2=15
only girls versus girls= 6 C 2=15
One girl versus one boy game=66-15-15=36
the fraction of the boys versus boys game=(total number of boys versus boys games)/ (total number of games)
=15/66
=5/22

This methodology can be used to solve similar problems with different number of boys and girls participating in the tournament

Ithink the answer is b