+0

0
42
4

Part A)   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. Of all the matches, how many of them have a left-handed player competing against a right-handed player?

Part B)   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. Of all the matches, how many of them have a left-handed player competing against a right-handed player?

Part C)   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.

May 24, 2021

#1
0

A) There are 50 matches.

B) There are 50 matches.

C) The fraction is 1/3.

May 24, 2021
#2
0

part a and part b are wrong :(

Guest May 24, 2021
#3
+1

10 people each have to play 9 other people = 90     BUT TWO people participate in each match

so  90/2 = 45 matches

May 24, 2021
#4
0

Part A is 25, part B is 10!

May 25, 2021