Finding number of ways of arranging people in teams in a tennis doubles game.

[cooldudeachyut]

Question - A mixed doubles tennis game is to be played between two teams. There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played.

My attempt - Well, the number of ways of arranging the men in team 1 is 4, and women should be 3(excluding the man's wife out of 4 females), and in team 2 the number of ways of arranging the men is 3 and women should be 3(including the wife of the man selected in team 1 and two other females), so according to me the answer should be (4*3*3*3)/2! {2! for not considering the team number as a factor}, which is 54 but the answer in my textbook is given 42. Where did I go wrong?

 

[pka]

Question - A mixed doubles tennis game is to be played between two teams. There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played.
the answer in my textbook is given 42. Where did I go wrong?

I agree with your text book. Unlike others (except Dennis) I think with reasonable numbers, students should work with models and avoid formuae.
Consider the players: \(\displaystyle F = \left\{ {\alpha ,\beta ,\gamma ,\delta } \right\}\;\& \;M = \left\{ {a,b,g,d} \right\}\) where the Greek females are married the the corresponding Anglo male.

Here is a matrix of possible pairs, no married couple.
\(\displaystyle \begin{array}{*{20}{c}}{(\alpha ,b)}&{(\alpha ,g)}&{(\alpha ,d)}\\{(\beta ,a)}&{(\beta ,g)}&{(\beta ,d)}\\
{(\gamma ,a)}&{(\gamma ,b)}&{(\gamma ,d)}\\{(\delta ,a)}&{(\delta ,b)}&{(\delta ,g)}\end{array}\) there are twelve.

There are three ways to select an alpha couple and then seven ways to select an opposing team?
Please explain why?!

Then there are two ways left to form the other match. WHY?

 

[cooldudeachyut]

There are three ways to select an alpha couple and then seven ways to select an opposing team?
Please explain why?!

Then there are two ways left to form the other match. WHY?

Same questions here.

 

[pka]

Same questions here.

Do you understand that we do not give instruction?
Do you understand why these are the only possible pairs:
Ab Ac Ad
Ba Bc Bd
Ca Cb Cd
Da Db Dc ? If you do not, there is no point in your continuing.

Because every player, all eight, will play in a match, there are three ways to pick a pair from row one. Say it is Ac.

In each of the remaining three eliminate any pair containing small c. WHY?

Then there are seven ways to select a second pair.

That leaves just two rows, BUT we have eliminate four people.
So how may ways can we complete the other match?

So \(\displaystyle 3\cdot 7\cdot ~?~=~?\)

 

[cooldudeachyut]

Do you understand that we do not give instruction?
Do you understand why these are the only possible pairs:
Ab Ac Ad
Ba Bc Bd
Ca Cb Cd
Da Db Dc ? If you do not, there is no point in your continuing.

Because every player, all eight, will play in a match, there are three ways to pick a pair from row one. Say it is Ac.

In each of the remaining three eliminate any pair containing small c. WHY?

Then there are seven ways to select a second pair.

That leaves just two rows, BUT we have eliminate four people.
So how may ways can we complete the other match?

So \(\displaystyle 3\cdot 7\cdot ~?~=~?\)

I did understand why those pairs are the ones taking part in the tournament. What I missed, was the obvious stuff and what you meant by alpha couple(English is not my native language). However, (forgive me if I appear to be a complete stupid) I still can't understand why there are only 2 rows left for arranging other matches. Shouldn't it be 4? Also, what mistakes did I make in my original attempt?

 

[pka]

I still can't understand why there are only 2 rows left for arranging other matches. Shouldn't it be 4? Also, what mistakes did I make in my original attempt?

Even though I do not care for the capitals, we will use
Ab Ac Ad
Ba Bc Bd
Ca Cb Cd
Da Db Dc as our model.

Lets make an example. We pick Ad. Because A & d are eliminated we can only choose from Be, Bc, Ca, Cb, Da, Db Dc.

This time we pick Bc. Because B & c are eliminated we can only choose from Ca, Cb, Da, Db.

Lastly picking Cb forces us to choose Da; picking Ca forces us to choose Db. That is two ways.