I think you've all got it wrong (n* (n-1))/2</p> (16 * 15)/2 = 120 different ties that could be played</p> That's means there are 120 different home or away combination of fixtures that Barnsley and the other teams could be playing.</p> I believe that paul d and jay have come up with the number of different cominations of orders that the balls come out of the bag. For example Barnsley vs Chelase is counted seperately depending upon whether it is the 1st, 2nd, 3rd, etc draw that comes out of the bag</p> With my calcualtion (if I've got it right) I'm saying Barnsely vs Chelsea is one possible fixture, and Chelsea vs Barnsley is another ... but the order of selection is not relevant and so I haven't accounted for it.</p> Dunno I think it's Coxter Groups and not Combinatorics you want to be looking at</p> </p> </p>
not Impossible it's the fact that Barnsley eg can have 14 possible games . Each of those 14 games can then combine with the 14 games (minus the overlaps) that each of the other 7 teams can have
haven't we have taken into account that eg AvB CvD EvF GvH is the same as eg AvB CvD GvH EvF when they are drawn in order
Nope, it doesn't matter when they come out of the bag Using your formula if there were only 4 teams the answer would be 6. Using my formula, the correct formula, the answer would be 12 and I can easily prove that. Barnsley, Man Utd, Chelsea and Bristol Rovers are in the draw. So the possibilities are 1Barnsley v Man Utd Chelsea v Bristol Rovers 2Barnsley v Man Utd Bristol Rovers v Chelsea 3Man Utd v Barnsley Chelsea v Bristol Rovers 4Man Utd v Barnsley Bristol Rovers v Chelsea 5Barnsley v Chelsea Man Utd v Bristol Rovers 6Barnsley v Chelsea Bristol Rovers v Man Utd 7Chelsea v Barnsley Man Utd v Bristol Rovers 8Chelsea v Barnsley Bristol Rovers v Man Utd 9Barnsley v Bristol Rovers Man Utd v Chelsea 10Barnsley v Bristol Rovers Chelsea v Man Utd 11Bristol Rovers v Barnsley Chelsea v Man Utd 12Bristol Rovers v Barnsley Man Utd v Chelsea See, 12 combinations. Easy innit.
I kniw I've screwed up on something 16 teams for a start.</p> I wasn't sure baout the division by two not accounting for home and away pairs.</p> I need to go away and think about it, but the figure you gave seems too high.</p> </p> What was the forumla you used to calculate 12?</p>
OK now I get it My calculation gives the number of pairs so in a league of 8 teams you've got (8 x 7)/2 = 28 pairs</p> That means 56 fixtures (home and away) in a league of 8 teams ... so there are 56 possible FIXTURES</p> However, you are right the number of permutations for the draw as in the different number of ways the draw could come out, and not the number of fixtures that could come out are the way you calculated</p> I was wrong, I'd misunderstood things. I always hated combination and permutation problems, I hate them even more now!!</p>