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

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A mixed doubles tennis game is to be played between two team [#permalink] New post 31 Aug 2012, 11:36
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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.

A. 12
B. 21
C. 36
D. 42
E. 46
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 02 Sep 2012, 23:05
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Spaniard wrote:
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.

A. 12
B. 21
C. 36
D. 42
E. 46


Let's denote the four couples by (A,a), (B,b), (C,c), and (D,d) where A,B,C,D are the husbands and a,b,c,d are the wives.
For a games, let's choose first the husbands. We have 4C2=4*3/2=6 possibilities. Now we know that the final number of games will be a multiple of 6, so we are down to two choices, C and D.

Once husbands were chosen, say A and B, let's count how many possibilities we have to choose their partners under the given restrictions.
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.
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Last edited by EvaJager on 05 Sep 2012, 08:36, edited 1 time in total.
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 05 Sep 2012, 07:26
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)
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 05 Sep 2012, 08:25
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euphrosyne wrote:
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
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PnC: A mixed doubles tennis game is to be played between [#permalink] New post 20 Oct 2012, 02:48
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.
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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 20 Oct 2012, 03:42
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avaneeshvyas wrote:
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.


IT is easy to find the number of games with married couples.

One married couple only:
Select one married couple out of 4 in 4C1 ways.
Select one male for the other team in 3 ways and one non-wife female in 2 ways.
Number of games with only one married couple = 4*3*2 = 24

Both married couples
Select 2 married couples in 4C2 = 6 ways

Number of games in which atleast there will be one couple = 24+6 = 30

Total number of games = (4*4 * 3*3)/2 = 72
Select team 1 in 4*4 ways and team 2 in 3*3 ways. Divide by 2 because you don't want to arrange the teams in team 1 and team 2. They are just 2 teams.

So in 72 - 30 = 42 games, there will be no married couple.
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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 20 Oct 2012, 04:06
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avaneeshvyas wrote:
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.


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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 31 Jul 2013, 06:01
VeritasPrepKarishma wrote:
avaneeshvyas wrote:
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.


IT is easy to find the number of games with married couples.

One married couple only:
Select one married couple out of 4 in 4C1 ways.
Select one male for the other team in 3 ways and one non-wife female in 2 ways.
Number of games with only one married couple = 4*3*2 = 24

Both married couples
Select 2 married couples in 4C2 = 6 ways

Number of games in which atleast there will be one couple = 24+6 = 30

Total number of games = (4*4 * 3*3)/2 = 72
Select team 1 in 4*4 ways and team 2 in 3*3 ways. Divide by 2 because you don't want to arrange the teams in team 1 and team 2. They are just 2 teams.

So in 72 - 30 = 42 games, there will be no married couple.



Can you please explain what the question is demanding, I have difficulty understanding the question correctly?

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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 31 Jul 2013, 21:04
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VeritasPrepKarishma wrote:
avaneeshvyas wrote:
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.


IT is easy to find the number of games with married couples.

One married couple only:
Select one married couple out of 4 in 4C1 ways.
Select one male for the other team in 3 ways and one non-wife female in 2 ways.
Number of games with only one married couple = 4*3*2 = 24

Both married couples
Select 2 married couples in 4C2 = 6 ways

Number of games in which atleast there will be one couple = 24+6 = 30

Total number of games = (4*4 * 3*3)/2 = 72
Select team 1 in 4*4 ways and team 2 in 3*3 ways. Divide by 2 because you don't want to arrange the teams in team 1 and team 2. They are just 2 teams.

So in 72 - 30 = 42 games, there will be no married couple.



Can you please explain what the question is demanding, I have difficulty understanding the question correctly?


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?
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 01 Aug 2013, 02:04
EvaJager wrote:
Spaniard wrote:
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.

A. 12
B. 21
C. 36
D. 42
E. 46


Let's denote the four couples by (A,a), (B,b), (C,c), and (D,d) where A,B,C,D are the husbands and a,b,c,d are the wives.
For a games, let's choose first the husbands. We have 4C2=4*3/2=6 possibilities. Now we know that the final number of games will be a multiple of 6, so we are down to two choices, C and D.

Once husbands were chosen, say A and B, let's count how many possibilities we have to choose their partners under the given restrictions.
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.



I have difficulty understanding this part-
we have 1 + 4 + 2 = 7 possibilities to choose their partners for the game.
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 02 Aug 2013, 22:41
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.
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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 04 Aug 2013, 03:53
VeritasPrepKarishma wrote:
avaneeshvyas wrote:
A mixed doubles tennis game is to be played between two teams(Each team consists of one male and one female). There are 4 married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

a)12
b)21
c)36
d)42
e)60

Detailed solution with brief description of each combination required.


IT is easy to find the number of games with married couples.

One married couple only:
Select one married couple out of 4 in 4C1 ways.
Select one male for the other team in 3 ways and one non-wife female in 2 ways.
Number of games with only one married couple = 4*3*2 = 24

Both married couples
Select 2 married couples in 4C2 = 6 ways

Number of games in which atleast there will be one couple = 24+6 = 30

Total number of games = (4*4 * 3*3)/2 = 72
Select team 1 in 4*4 ways and team 2 in 3*3 ways. Divide by 2 because you don't want to arrange the teams in team 1 and team 2. They are just 2 teams.

So in 72 - 30 = 42 games, there will be no married couple.

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 ??
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 21 Aug 2013, 03:12
Please check and provide feedback for my solution :

First we will find possible number of teams then we can find total number of games .

M1 W1 --- With M1 --3 Pairs ..so three teams
M2 W2
M3 W3
M4 W4 ----- So total 12 teams .

Now we have to find out total games = 12C2 - 4(3!) = 42
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Re: PnC: A mixed doubles tennis game is to be played between [#permalink] New post 21 Aug 2013, 21:19
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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.
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 21 Aug 2013, 21:34
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hsb91 wrote:
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.


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

Quote:
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 )


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.

Quote:
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


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
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 02 Oct 2013, 07:38
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.
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Re: A mixed doubles tennis game is to be played between two team [#permalink] New post 02 Oct 2013, 20:34
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bsahil wrote:
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
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Re: A mixed doubles tennis game is to be played between two team   [#permalink] 02 Oct 2013, 20:34
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