Combinatorics: tennis game with 8 people

member 587159

Homework Statement

8 friends are playing a tennis game together. How many different doubles games of tennis can they play?

Combinations

The Attempt at a Solution

Well, I solved this problem by saying: we choose a group 4 people from 8 to play, so order is not important, this is C(8,4). If those 4 people are A,B,C,D, they can play together in 3 different ways: AB/CD; AC/BD; AD/BC, so the solution is: 3C(8,4) = 210.

I tried doing this in another way, but I suppose I am doublecounting something. This was my approach:

For one team, we choose 2 people out of 8. For the other team, we choose 2 people out of 6 (since 2 out of 8 players are already taken). So, the answer would be: C(8,2)*C(6,2). Now, I am pretty sure that I have doublecounted, but I don't know exactly what it is I have doublecounted.

It would be great of someone could help me!

mfb
Mentor
For one team, we choose 2 people out of 8. For the other team, we choose 2 people out of 6 (since 2 out of 8 players are already taken). So, the answer would be: C(8,2)*C(6,2). Now, I am pretty sure that I have doublecounted, but I don't know exactly what it is I have doublecounted.
You counted (AB play against CD) and (CD play against AB) as separate events.

member 587159
Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

8 friends are playing a tennis game together. How many different doubles games of tennis can they play?

Combinations

The Attempt at a Solution

Well, I solved this problem by saying: we choose a group 4 people from 8 to play, so order is not important, this is C(8,4). If those 4 people are A,B,C,D, they can play together in 3 different ways: AB/CD; AC/BD; AD/BC, so the solution is: 3C(8,4) = 210.

I tried doing this in another way, but I suppose I am doublecounting something. This was my approach:

For one team, we choose 2 people out of 8. For the other team, we choose 2 people out of 6 (since 2 out of 8 players are already taken). So, the answer would be: C(8,2)*C(6,2). Now, I am pretty sure that I have doublecounted, but I don't know exactly what it is I have doublecounted.

It would be great of someone could help me!

There are C(8,4) ways of choosing a group of 4 to play doubles; for each such group of 4 there are C(4,2) ways of forming two teams of two.

member 587159
You counted (AB play against CD) and (CD play against AB) as separate events.

I still don't get it. Where exactly do I do this?

PeroK
Homework Helper
Gold Member
2020 Award
There are C(8,4) ways of choosing a group of 4 to play doubles; for each such group of 4 there are C(4,2) ways of forming two teams of two.

Not quite. That double counts the ways of forming 2 teams of 2.

PeroK
Homework Helper
Gold Member
2020 Award
I still don't get it. Where exactly do I do this?

C(8, 2) includes AB as a team; given AB is the first team, C(6, 2) includes CD as a team.

C(8,2) includes CD as a team; given CD is the first team,C(6,2) includes AB as a team.

member 587159 and mfb
member 587159
C(8, 2) includes AB as a team; given AB is the first team, C(6, 2) includes CD as a team.

C(8,2) includes CD as a team; given CD is the first team,C(6,2) includes AB as a team.

Thanks a lot!

Ray Vickson
Homework Helper
Dearly Missed
Not quite. That double counts the ways of forming 2 teams of 2.

If the group is ABCD we can choose one of the pairs to be AB or AC or AD or BC or BD or CD. Whenever we make such a choice, the remaining two form the other pair.
Not quite. That double counts the ways of forming 2 teams of 2.

Of course, I missed that. If the group is ABCD there are only three pairs that A can belong to, and he/she must belong to some pair.

haruspex