There are 13 teams in a tournament. Each team is to play with each other only once. What is the minimum number of days can they all play without any team playing more than one game a day

Team one must play against 12 different temas = > 12 days

Team two, already played with team one, then only has to play 11 more days

Team three, already played with teams one and two, then there are 10 dates left.

So on, the result is 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 78 days.

You can also find that the number of different games is 13*12 / 2! = 78

Also, you can do Combinations of 13 taken 2 at a time: 13C2 = 13! / [ (2!) (11!) ] = 13*12*11! / (2!*11!) = 13*12/2 = 78.