1. Split the total population of different ways into two groups: Group 1) duplicate rounds only ... for example ... A-M, B-N, C-O for two rounds, and Group 2) non-duplicate rounds only ... for example ... A-M, B-N, C-O for round 1 then A-M, B-O, C-N for round 2. 2. Group 1 population = 2 distinct ways to have duplicate rounds only, ignoring order. Looking at just one of those distinct ways and factoring-in order gives us 6! = 720 ways, but each pair of duplicate rounds (all 3 of them) are repeated twice in that 720 population, so we need to divide by 2 for all 3 pairs of duplicates. So, in general, this gives us 6! / (2! * 2! * 2!) * 2 (multiply by two because, remember, there are 2 distinct ways to have duplicate rounds only) = 180 unique ways the match can be scheduled using duplicate rounds only. 3. Group 2 population = 1 distinct way to have non-duplicate rounds only, ignoring order. Factoring-in order gives us 6! = 720 unique ways the match can be scheduled using non-duplicate rounds only. 4. Group 1 + Group 2 = Final answer, --> 180 + 720 = 900

Is this real?

What kind of screen are you using to write on with your finger?