maybe I'm not so skilled in combinative formulas but for me this is the way: for even N the quantity of pairs = (N-1)*(N-3)*...*(N-n), where n = N-1. So that for 8 players it's 7*5*3*1=105. For N=10 it's 9*7*5*3*1 = 945

P.S. I should explain how I reached that. I draw a picture with 4 lines then investigated those combinations, it was 3. Why 3, cause if we chosen 1 pair only 1 pair left. How many first pairs I could chose - 3 (N-1).

OK, how about 6? I chosen 1 pair and what I saw, right, 4 lines. I had already seen them and knew that they consisted 3 possible pairs. How many first pairs I might chose, correct, 5 (N-1). So that we had (N-5)*(N-3)*(N-1). I kept reasoning and found the regularity and wrote it in the opposite way - see above

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All this reasoning took around 1 minute...