tennis maths probability (1 Viewer)

[mathsfreak]

29a) how many doubles tennis games are possible, given a group of four players?
b) In how many ways can two games of doubles tennis be arranged, given a group of eight players
c) six married couples are to play in a three games of double tennis. Find how many ways the pairings can be arranged if
i) there are no restrictions, ii) each games is to be a game of mixed doubles


im just really confused with this concept, i get how there are three ways for a) because those games are the only possible ones
but for 8 players, it too hard to count individually.


thanks in advance

 

[braintic]

29a) how many doubles tennis games are possible, given a group of four players?
b) In how many ways can two games of doubles tennis be arranged, given a group of eight players
c) six married couples are to play in a three games of double tennis. Find how many ways the pairings can be arranged if
i) there are no restrictions, ii) each games is to be a game of mixed doubles


im just really confused with this concept, i get how there are three ways for a) because those games are the only possible ones
but for 8 players, it too hard to count individually.


thanks in advance

(b)

Pick 4 players from 8 for one doubles game: 8C4

This must be divided by 2 because, for example, picking A,B,C,D for one game and leaving E,F,G,H behind for the second game is exactly the same as picking E,F,G,H and leaving A,B,C,D behind.

Then by part (a), each game can be arranged 3 ways.

So the total number of arrangements is (8C4 / 2) * 3 * 3 = 90720




(c) (Not sure what the married couples part is about - it doesn't get used)


(i) (I assume 'no restrictions' means you can mix up the men and women as you please)

(12C4) * (8C4) / (3!) * 3^3 = 155 925

See if you can work out the logic based on the previous question.


(ii)

First pretend it is a men's singles tournament with 3 games

(6C2) * (4C2) / (3!)
by the same logic as previously (there is no need to multiply by anything at the end, because once you pick the two players involved in a match then there is only one possible matchup involving those players)

Then match up each of the women with a man:
First woman has 6 possibilities, 2nd has 5, ....
So 6! ways.

So answer is (6C2) * (4C2) / (3!) * (6!) = 10800



Let me know whether these answers match up with the ones you have been given.

 

[mathsfreak]

yeh the answers are correct

however (8C4 / 2) * 3 * 3 = 315 which is the right answer (probably typed it in the calculator wrong?)

thank you soooooo very much