A group of 8 friends wants to play double tennis. How many

When we choose four groups, it is necessary to exclude order among groups, for example:

12,34,56,78 and 34,12,56,78 is the same option. (1,2,3,4,5,6,7,8 - people)

Can u explain the logic behind this formula?

\(C^8_2\) - choose first team

\(C^6_2\) - choose second team

\(C^4_2\) - choose third team

\(C^2_2\) - choose fourth team

\(P^4_4\) - exclude order among four teams.

Pretty good explanation Walker

for some reason i get 115..

i did as follows.. [8C2*6C2*4C2*2C2]/4!

i use 4! because suppose each player is a nd B..then we have counted that pair twice i.e ab, ba etc for each of the 4 slots..

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 .
All this reasoning took around 1 minute...