a tricky maths problem solving question. Interesting, but hard! I really need some help to work this out!!
A tennis club is running a mixed doubles tournament for families. the families enter one male and one female. they soon discover that this is arranged as a tounament with a twist (TWT), in that they NEVER PARTNER OR PLAY AGAINST THEIR OWN FAMILY MEMBER.
the TWT is arranged so that:
1. each player plays against every person of the same gender exactly once.
2. each player plays against every person of the opposite gender, except for his or her family member, exactly once.
3. each player partners every person of the opposite gender, except for his or her family member, exactly once.
using the notation M1 and F1 for the male and female from family 1, M2 and F2 for family 2 and so on. An example of an allowble match is M3F1 v M6F4.
QUESTION:find all TWTs for four families.
I have no idea how to start solving this quetsion! its very hard. Im trying to do it systematically and list out all the possibilites, but its very long!! I really need some help. How ive started doing it will take forever (im kind of doing it like a tree diagram, showing all the options but its really big!!) how big do you think the answer will be? does the same set of matches in a different order make a different tournament? or is it a copy? HELP ME!! anything is appreciated!!
Ohh.......! It is a very good question if you get any answer please forward to me also . My email i d is email@example.com
Indeed this is quite tricky but i think i cracked it, where did you get this, it looks like an excercise or homework or something. well you did say anything will do, the concept lies on the end of each additional notes (exactly once), and where did you get the example, its simple. whats on the right , you don't put on the left. And since its a four family one, it will be (1,2,3,4) thats it. Therefore: all TWTs for four families are ; 1. M1F2 VS F3M4
2. M2F3 VS M4F1
3. M3F4 VS F2M1
Thats it, if you can manage to find another combination, then do so, but i think you can't since according to the laws of subset( thats right , you see how simple this thing is) which is 2 to the n power, you wouldn't be to find any other pair combinations, they would be already use, if not, be on the other side. It took like 2mins or so to figure it out, i know its long since you send it but hey no one answered. i like these kind of stuff, do send more!