A mixed doubles tennis game is to be played between two team

Can you please explain your solution once more to me? According to my calculations its 36.
4C2(=6) to choose 2 husbands out of the 4. Now for each husband chosen there are 3 wives that can be paired. So-6*3*2 (cause there are two other husband- wife mixed double combination possible.)
so- 36 answer (c)

After we have chosen the pair of husbands, A and B:
We can choose a and b and we have to pair them as (A,b) and (B,a) - 1 possibility.
We can choose one of the wives, a or b, but we have to pair her with the other husband and in addition, we have to choose another partner for the second husband.
This we can do in 2*2 = 4 ways, as there are two possibilities to choose from a and b, then we have 2 possibilities to choose the other wife, c or d - 4 possibilities
Finally, we can choose the other two wives, c and d, and we have two possibilities to team them up with the men, (A,c), (B,d) or (A,d), (B,c) - 2 possibilities.
In conclusion, for every pair of husbands, we have 1 + 4 + 2 = 7 possibilities to choose their partners for the game.
Total number of possibilities 6 * 7 = 42.

Answer D.

Now for each husband chosen there are 3 wives that can be paired. So-6*3*2
There are no 6 husbands, but only 4. You should not consider husbands alone.
You should count the possibilities of choosing the wives per chosen pair of husbands, otherwise you cannot keep up with repetitions or you can miss out some possibilities. 6 represents the number of pairs of husbands. Then the number of 3 wives is not correct, as you can see from the above, all four wives can be candidates as partners, depending whom each plays.

If you write down all the possibilities for the pair of husbands A and B, you will get a total of 7 (following the steps described above):
(A,b) (B,a) - 1
(A,b) (B,c); (A,b) (B,d); (A,c) (B,a); (A,d) (B,a) - 4
(A,c) (B,d); (A,d) (B,c) - 2

Or in other words:
1 - two wives stay, but each has to play with the other husband
2 - one wife stays and plays with the other husband, and second husband receives one of the other two wives as a partner
3 - none of the wives stays, the remaining two wives pair up with the already chosen husbands

The question says that a mixed double match is to be played. You need two teams - each team consisting of one man and one woman - for the match.
You have 4 married couples who would be willing to play - 4 men and 4 women. So you have to choose 2 men and 2 women such that they form two teams of one man one woman each in which the man and woman are not married.
So say, you choose Man1 and Man3 to play. Now you have to choose 2 women. With Man1, you cannot have Woman1 but you can have any of the other 3 women.

So the two teams could look like this:
Man1-Woman2 and Man3-Woman1
or
Man1-Woman3 and Man3-Woman2
or
Man1-Woman3 and Man3-Woman4
etc

What you should NOT have is something like this:
Man1-Woman1 and Man3-Woman2
or
Man1-Woman2 and Man3-Woman3
or
Man1-Woman1 and Man3-Woman3

So the question is : In how many different ways can you make such a team?

My approach is similar to Karishma but i'm not able to arrive at correct choice.

For team1 we have two places to be filled - 1 for man and 1 for woman... 4*4 = 16 ( without any restrictions )
For team 2, 3*3 = 9 (no restriction)

For a match 16*9/2! = 72 total ways ( T1 = 16 ways and T2 = 9 ways. Divide by 2! as order/arrangement not required )

Now consider case where 1 couple is playing.
Team 1 -- for man 4 ways and for woman only 1 possibility. Thus a total of 4 ways.
Team 2 -- for man 3 ways and for woman 2 ways. Thus 6 ways.
Now, for a match 4*6/2! = 12 or simply 24( as done by Karishma )

Case2, two couples.
Team 1= man in 4 ways and woman in 1 way = 4 ways
Team 2 = man in 3 ways and woman in 1 way = 3 ways
For a match 4*3/2! or 4*3 ??

Please clarify.

Karishma, please reply to my post and point out my mistake in the approach i have used.. And is no. of ways of formation of two teams with the given restrictions same as the no. of matches possible ??

'Maximum number of games' implies 'in how many distinct ways can you make the teams'. e.g. (M1, W2 and M2, W3) OR (M1, W3 and M3, W4) etc. This is implied by the context; though if you take the question literally, it makes little sense.

There are some little concepts here which make a big difference.



The number of ways here is 24, not 12. You will not divide by 2 here. The teams are distinct. One has a couple and the other non couple. You select a couple in 4 ways and then the non couple in 3*2 = 6 ways. Total 24 ways. There is no double counting here.



Here, you do need to divide by 2 because there is double counting. e.g. Team 1: M1, W1, Team 2: M2, W2 OR Team 1: M2, W2, Team 2: M1, W1.
Or you can just use 4C2 i.e. select 2 of the 4 couples.
4C2 = 4*3/2 = 6 ways

Hello,

I started with following way but stuck in the end.
First we make number of teams possible.:
4C1*3C1 = 12 teams are possible.

Number of matches = 12C2= 66

Please help me further.

This approach is incorrect.

You can make 12 distinct teams - that's fine. They will look like this:
AB'
AC'
AD'
BA'
BC'
BD'
CA'
... etc
Now can you pick any two out of these and have a game? Say you pick AB' and AC'. Can A play as the male member on both teams in a game?

You have to think in terms of a game instead.
Say, you select 2 male members out of a total of 4 in 4C2 ways. Say you select A and B.
Now for one male member, say A, you can select a partner in 3 ways (B', C' and D'). The problem is that if you select B', you have 3 options for B's partner (A', C' and D'). IF you select C' or D' for A, you have only 2 options for B (A' and C'/D' whoever is left). So you take two cases:

Select B' for A --> 4C2* 1 * 3 = 18

Select other than B's wife for A --> 4C2 *2*2 = 24

Total number of ways = 42

I solved it in a different way and ended up getting wrong answer. Still I am not able to find a mistake in my method. It will be great if you could help me finding my mistake.

the ways in which teams can be formed = 4C1 * 4C1 (one male out of four and one female out of four) = 16, but this includes the married couples in one team. So number of ways in which teams can be formed = 16-4 =12

number of games among 12 teams = 6------> 6 winner teams

number of games amaong 6 teams = 3-------> 3 winner teams

number of games among 3 teams = 1--------> only two teams can play a match, number of winner =1

last game between 2 teams ------> final winner

so total number of games played = 6+3+1+1 = 11

please help me finding my mistake.

The wording of the question is a little off.

"What is the maximum number of games that can be played?"
actually means
"In how many different ways can you make the two teams?"

Only one game is to be played. You need two 2-people teams for that. A married couple should not be a team. In how many different ways can you make the two teams?

Now review your solution.

My approach is

My interpretation is in no game a husband & wife should play together either in same team or opposite team ( SO 12 is correct answer if that was the question)

When husband & wife can play a game but in opposite teams, I think we can take the following approach.

a. For each game we require team A & Team B

for Team A we have two spots: Male ___ and ___ Female ( Mixed doubles)

Here first Husbands can be selected in 4 ways x 3 ways one of the female partners: so team A can be selected in 12 ways. ( to remove repetitions we need to divide by 2) Thus = 6 ways.

Now Come to team B:

Here we need one male ____ and ____ one female.

So Male partner in B team we have 3 ways (it can be the husband of the already chosen female) x 2 ways for the female partner. ( as his wife cannot be chosen here in team B) TO remove repetition we need to divide by 2. Thus we have 3 ways for team B.

Now all we need is to select both A & B teams SO totally we need : 6 x 3 games = 18 games.


Alternatively:

Total games:
8C4 = 70 games

Not out of these the following games are not valid:

1. Teams in which husband & wife are together in either of the team

4 (1 couple out of 4 couples for team A can be chosen in 4 ways) x 6C2 (Opposite team can be couple or any general) = 4 ways x 15

But a game between A & B is same as that between B & A so we need to divide by 2. Thus we have 60/2 = 30 invalid games.

Thus, total games are = 70-30= 40 games all together.

Can anyone explain where I am going wrong??

Also can you explain why 8C4 is wrong when choosing 4 people out of 8.

Essentially when you do 4 x 4 x 3 x 3 /2 you are doing the same.

Dear ankushbagwale

Here's why 8C4 is wrong:

8C4 refers to choosing 4 people out of 8 people. Had the question been: 'there are 8 people and a team of 4 is to be formed out of them. In how many ways can you form this team?' Then, the answer is 8C4.

Note that here we are just selecting one bunch of 4 people. And, we are not organizing them in any way at all.

Now consider this case: 'There are 8 men and 2 doubles teams are to be formed out of these 8 men. In how many ways can you form these teams?'

Will your answer still be 8C4?

To be sure, think about this: are you simply selecting one bunch of 4 people here?

No, right? There are 2 ways in which you can think about what you are doing here:

Way 1 to think:
1. You are selecting 2 men out of 8 men for Team 1. In how many ways can you do this? 8C2
2. From the remaining 6 men, you select 2 men for Team 2. In how many ways can you do this? 6C2

So, the number of ways to select 2 men for Team 1 AND 2 men for Team 2 = 8C2*6C2 = 420

Way 2 to think:
1. You are selecting a bunch of 4 players out of the 8 men. In how many ways can you do this? 8C4
2. You now distribute these 4 players in teams T1 and T2. For example: Let the 2 players of T1 be P1 and P2. We have 4 choices for P1 and 3 choices for P2. And since order doesn't matter, the number of ways in which Team T1 can be made out of the 4 players = 4*3/2 = 6. Once team T1 is made, the remaining 2 players automatically make Team T2.

So, the number of ways to select 4 players AND then distribute them in 2 teams = 8C4*6 = 70*6 = 420


Finally, let's come to the case at hand: 'There are 4 men and 4 women and 2 MIXED doubles teams are to be formed out of these 8 people. In how many ways can you form these teams?'

Think about what you need to do here:

i) Select 2 men players out of the 4 men. This can be done in 4C2 ways
ii) Select 2 women players out of the 4 women. This can be done in 4C2 ways
iii) Distribute these 2 player men and 2 player women in teams T1 and Team T2. This can be done in 2 ways (W1M1 v/s W2M2 or W1M2 v/s W2M1)

So, the number of ways to select 2 men and 2 women players for 2 MIXED DOUBLES teams = 4C2*4C2*2 = 72

I hope the distinction between the different cases is clear now. Once you've understood this, you'll also be able to see why the highlightedexpression in your solution above is wrong. (Hint: 6C2 would have been the number of ways to select Team 2 had gender not mattered in the team composition; this number also includes men-men and women-women selections)

Once you correct the highlighted expression, we can then discuss the brown part.

Do let me know if something remains unclear.

Best Regards

Japinder

Dear Japinder,

Many thanks for the detailed explanation & the clarity with which you have conveyed the microscopic nuances of the fundamental principles of counting.

If I am correct & able to get what you have conveyed, I think the error in the highlighted part is :

6C2 would have been correct had the question allowed any team ( meaning all mens & all womens team also permitted. That's not the case here.)

The opposite team ( either a couple or a non couple would render the game invalid as already team 1 formed with a couple)

So the number of ways : ( # of ways of a couple OR a non couple. )

1. # of ways for a couple: 1 out of remaining 3 couples = 3 ways

2. # of ways for a non couple: for first position : we have 6 places x we cannot have a partner of the couple so we we have 4 choices = 24.
Here order does not matter hence ( 24 /2 ) = 12 ways

Hence the total ways for selecting the team 2 ( Either a couple OR a non couple ) = 3 + 12 = 15

Thus the total number of invalid games = 15 x 4 = 60

Now we had total games : 420 - Invalid games 60= 360 valid games ( out of which 4 are common so 360/4 = 90 valid games??)

There is no option of this sort. I am sure I am getting somewhere with this analysis but drifting away from the correct response.

Not so. The question says that there will be two teams. No team is to consist of a husband and his wife. So a team should not have a married couple. It doesn't say that there should be no married couple across teams.


There is no repetition here. For team A, you select a guy in 4 ways and a girl in 3 ways. How can you double count here? We are selecting the two people from different groups. Team A can be selected in 12 ways.

Double counting happens if you select from the same group of people. If you were to select 2 people (any two people) from 4 people, you can select the first one in 4 ways and the second one in 3 ways. Now you have to divide by 2 because there is double counting. The first one could be A and second one B. In another case, first one could be B and second one A. Here you have to select from the same group.

The repetition happens only after you select both team A and team B because actually there are just two teams. So Team A: Ha, Wc and Team B: Hd, Wa is the same as Team A: Hd, Wa and Team B: Ha, Wc
So that is when you will need to divide by 2.


Now there is a problem here. If the wife of the guy we select now was selected in team A, we have 3 options to select his female partner. If she wasn't selected in team A, we have 2 options to select his female partner. Say, if we selected Ha and Wc as team A. If Hc is selected in team B, you have 3 options for Wife here (Wa, Wb or Wd). If Hb is selected in team B instead, then you have 2 options for wife (Wa or Wd).
That's where it all goes wrong. Hence, it is far easier to select couples than to make teams without couples. Still, now that we have come this far, let's wrap this up:

Here you will take 2 cases:
Husband's wife already in team A: Team can be selected in 1*3 ways
Husband's wife not in team A: Team can be selected in 2*2 = 4 ways
Total 7 ways

Total ways of selecting team A and team B = 12*7 = 84 ways
But we are double counting. There is no team A and B. They are all just teams so 84/2 = 42 ways.

Hi Karishma VeritasKarishma
I'm just confused . Are the teams supposed to be not married a man and a woman ?
did your solution based on that , because I solved it as any 2 not married people
can you please let me know why my approch is wrong :

let the teams : A a , B b , C c , D d
let's choose 4 people of those 8 in 8C4 ways = 70 different sets , each has 4 people
now , each set can perform the games in 3 ways , for example , let's say that the set is "a b c d ",
they can perfrom the game in 3 ways (ab * cd ) (ac * bd ) (ad * bc)
so , the total number of games is 70 * 3 = 210

now let's find the number of games which contains a married team:
one married team VS not married team + 2 married teams
we can choose a married team in 4 ways , for one side of the game
for the other side , 6 people left , so we need 2 of them and we exclude the 3 married ones ,
6C2 -3 = 12
12 *4 = 48

2 married teams : can play together in 3 different ways , so the number of games which contains at least married team
is 48 +3 = 51

so the answer must be 210 - 51 = 159

This is what the questions says: "There are 4 married couples. No team is to consist of a husband and his wife."

So all 8 people are married people.
Ah, Aw
Bh, Bw
Ch, Cw
Dh, Dw

A mixed doubles match will need two teams, each team having one male and one female player. The question says that you cannot have a husband and his wife so you can't have Ah and Aw together as one team.
The teams could be Ah, Cw and Bh, Aw etc.

Does this help clarify?