[sub]128[/sub]N[sub]6[/sub] = 5,423,611,200
[sub]6[/sub]N[sub]3[/sub]/2 = 10
[sub]128[/sub]N[sub]6[/sub]*[sub]6[/sub]N[sub]3[/sub]/2 = 54,236,112,000 (possible matches)
So for a 40-match tournament,
[sub]54,236,112,000[/sub]N[sub]40[/sub] = 2.88x10381 (possible different “tournaments”, not counting the order that mathces are played)*
*Of course, the vast majority of these “tournaments” are patently ridiculous - like one team playing all 40 matches and another not playing at all. With just a bit of smarts, the number could be reduced greatly. And, unless you are designing an algorithm to examine all possible tournaments and pick the “best” one, you don’t have to deal with this search space
**