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Re: A mixed doubles tennis game is to be played between two team
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27 May 2015, 08:20
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.



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Re: A mixed doubles tennis game is to be played between two team
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27 May 2015, 20:37
ankushbagwale wrote: 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)
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. ankushbagwale wrote: 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.
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. ankushbagwale wrote: 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 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.
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Re: A mixed doubles tennis game is to be played between two team
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27 May 2015, 23:59
VeritasPrepKarishma wrote: ankushbagwale wrote: 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)
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. This is the part where I have faltered. I am glad to know this hurry in understanding the problem & drawing unnecessary assumptions.
VeritasPrepKarishma wrote: 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.
[/quote] Many thanks for clarifications on my approach & highlighting fundamental issues in the approach. I would like to know an easier / intuitive approach to this question. Btw why the negation method won't work here is something I am interested in.



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Re: A mixed doubles tennis game is to be played between two team
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28 May 2015, 00:17
I have used the negation method in a post above  I found the number of ways in which you can have at least one couple in the team and then subtracted that out of the total. Check it out.
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Re: A mixed doubles tennis game is to be played between two team
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28 May 2015, 05:15
ankushbagwale wrote: 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.) That's correct ankushbagwale ankushbagwale wrote: 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
Multiple mistakes here: In the red part, you are taking the OR case only for the selection of Team 2. The correct way to look at it will be: Invalid teams are formed with: ( A couple in Team 1 AND a couple in Team 2) OR (A couple in Team 1 AND a noncouple in Team 2)A couple in Team 1 AND a couple in Team 2 can be selected in 4C1*3C1/2 = 6 ways. We divide by 2 because order doesn't matter here: H1W1 v/s H2W2 is the same match as H2W2 v/s H1W1. (Alternatively, you can look at this as selecting 2 couples out of the 4 couples, 4C2 = 6) A couple in Team 1 AND a noncouple in Team 2 can be selected in 4C1 (numbers of ways to select 1 couple out of 4 couples)*3C1 (number of ways to select one of the 3 remaining men) * 2C1 (number of ways to select one of the 2 women that are not the wife of the selected man) = 24So, number of invalid teams = 6 + 24 = 30Now look at the highlighted part in your solution: When you write the number of ways to form a noncouple Team 2 as 6*4/2, what you are saying is that if you've selected H2 as a player for Team 2, then only W2 (wife of H2) is not allowed to be the other player of Team 2 apart from W2, any of the remaining 4 people can be a part of Team 2. But that is not true! H2H3 or H2H4 are not valid combinations (because these are manman combinations). ankushbagwale wrote: 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. The total number of combinations possible is 72, not 420 (look at the posts above in case of any doubt) So, number of valid games = 72  30 = 42 We ended up having quite a rich discussion. I hope it was useful! Best Regards Japinder
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Re: A mixed doubles tennis game is to be played between two team
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05 May 2017, 16:14
Tough question.
Here's my method. There will always be two men, one on each team. There are 4C2 = 6 ways to do that. Now to deal with the restrictions. The numbers aren't huge so let's see what ONE of those combinations could look like to see if we can apply the rule of product to the remaining 6. Men = {A,B,C,D}, Women ={a,b,c,d}
(A,b) (B,a) (B,c) (B,d) (A,c) (B,a) (B,d) (A,d) (B,a,) (B,c)
Seven possibilities per selection of competing men. 6*7 = 42.



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Re: A mixed doubles tennis game is to be played between two team
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16 Mar 2019, 16:15
VeritasKarishma 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 nonwife 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. 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
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Re: A mixed doubles tennis game is to be played between two team
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16 Mar 2019, 21:59
foryearss wrote: VeritasKarishma 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 nonwife 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. 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?
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A mixed doubles tennis game is to be played between two team
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16 Mar 2019, 22:26
Thanks for your reply Karishma , but still , I don't understand why can't we consider "Ah Bh" as a team , 2 males or 2 femals the question didn't say that the team must be a husband and a wife ,it only sais that we can't but a husband and his wife together , right ?
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Re: A mixed doubles tennis game is to be played between two team
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16 Mar 2019, 23:11
Thanks a lot , I got it
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Re: A mixed doubles tennis game is to be played between two team
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17 Mar 2019, 12:39
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 The easier method is to subtract the couple pair from the total no. of possible pairs 1. Total no. of pairs possible: No. of ways of choosing two men = 4C2 = 6 No. of ways of choosing two women = 4C2 = 6 No. of ways of forming teams = 2 so total no. of pairs possible = 6 * 6 * 2 = 72 2. Total no. of couples team Only one team has couples in it: No. of ways choosing one couple = 4C1 = 4 No. of choosing one man and one nonwife woman or viceversa = 3C1 * 2C1 = 6 so total no. of pairs with only one couple = 4 * 6 = 24 Both teams have couple in them No. of ways of choosing two couples = 4C2 = 6 So, total no. of ways of choosing couples team = 24 + 6 = 30 so total no. of teams without having any couple in them = 72  30 = 42. Choice D




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