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 two left-handed players?
The number of such matches equals the number of ways that you can choose 2 left-handed person from 5 of them to play a match, which is \(\displaystyle\binom{5}{2} = 10\).