Some of the thoughts on the way to this result: First, how many ways are there to arrange the players in one particular set? If we number the eight spots on the two courts 1-8 and assign players to them you can say there are 8! = 40320 ways to assign the players for this set. But we don't really care to assign the left/right position of each player on each side of the net, or even which side of the net teams are on, or which of the 2 courts each foursome is on. So 8! overcounts by a factor of 16 for the left/right assigment, 4 for the side of net assignment, and 2 for the court assignment, i.e. we should divide the 40320 by 128, giving 315.

First set: 12 v 34, 56 v 78 Second set: 13 v 57, 24 v 68 Third set: 14 v 58, 23 v 67 Fourth set: 15 v 26, 37 v 48 Fifth set: 16 v 38, 25 v 47 Sixth set: 17 v 46, 28 v 35 Seventh set: 18 v 27, 36 v 45

Now there are binomial co-efficient (315 over 7), or 57,093,827,594,310 selections from these 315 set-lineups for a 7-set tournament (considering the order of the sets to not be relevant). Of these, apparently 720 meet the requirements (each as partner once, each as opponent twice), one of which I show above.