A group of 4 married couples want to play mixed tennis, but each married individual does not want to be on the same team as his/her spouse. How many possible games can be played between two teams?
(A) 12
(B) 21
(C) 36
(D) 42
(E) 46
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Mission2012
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lunarpower
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Maybe I'm misunderstanding something, because I get 156 possible games.
Here's how I figure it:
let's say the couples are AB JK PQ XY.
--
fact 1:
there are 24 possible teams.
* list them: AJ AK AP AQ AX AY BJ BK BP BQ BX BY JP JQ JX JY KP KQ KX KY PX PY QX QY.
* combinatorially: 8 choices for the first team member; 6 choices for the second (can't pick the spouse, so not 7).
if you just do 8 x 6, you're double-counting every team, so there are (8 x 6)/2 = 24 teams.
--
fact 2:
each of these 24 teams can play any of 13 opposing teams.
i don't see a decent combinatorial way here, so, let's just make a list for the teams that "AJ" could play:
BK BP BQ BX BY KP KQ KX KY PX PY QX QY
there's nothing special about "AJ", so every hypothetical team has 13 possible opposing teams.
--
ok, let's combine these facts.
24 x 13 games... but, then, you're double-counting each game. so, there should be (24 x 13)/2 = 12 x 13 = 156 possible matchups.
is this the exact wording of the original?
if so, what does the answer key say?
iiiiiiiiinteresting...
Here's how I figure it:
let's say the couples are AB JK PQ XY.
--
fact 1:
there are 24 possible teams.
* list them: AJ AK AP AQ AX AY BJ BK BP BQ BX BY JP JQ JX JY KP KQ KX KY PX PY QX QY.
* combinatorially: 8 choices for the first team member; 6 choices for the second (can't pick the spouse, so not 7).
if you just do 8 x 6, you're double-counting every team, so there are (8 x 6)/2 = 24 teams.
--
fact 2:
each of these 24 teams can play any of 13 opposing teams.
i don't see a decent combinatorial way here, so, let's just make a list for the teams that "AJ" could play:
BK BP BQ BX BY KP KQ KX KY PX PY QX QY
there's nothing special about "AJ", so every hypothetical team has 13 possible opposing teams.
--
ok, let's combine these facts.
24 x 13 games... but, then, you're double-counting each game. so, there should be (24 x 13)/2 = 12 x 13 = 156 possible matchups.
is this the exact wording of the original?
if so, what does the answer key say?
iiiiiiiiinteresting...
Ron has been teaching various standardized tests for 20 years.
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lunarpower
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by the way, in case anyone is stressing about the ostensible amount of work involved in "fact 2" there ... not really; i just took the list i made for fact 1 and crossed off all the teams with A or J on them. so, this problem can easily be done within a reasonable time, even by "brute force".
still... yeah. would like to see the answer key for this one. what's the source?
still... yeah. would like to see the answer key for this one. what's the source?
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
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On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Brent@GMATPrepNow
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Hey Ron,lunarpower wrote:
The question says "mixed" tennis, which means each team consists of 1 man and 1 woman. So, several of your 24 possible teams aren't possible.
Of course, the GMAT would never require test-takers to know what mixed tennis means.
Cheers,
Brent
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Brent@GMATPrepNow
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Here's one approach:Mission2012 wrote:A group of 4 married couples want to play mixed tennis, but each married individual does not want to be on the same team as his/her spouse. How many possible games can be played between two teams?
(A) 12
(B) 21
(C) 36
(D) 42
(E) 46
First, select 2 men. These two men will be on opposite teams.
Since the order of the selected men does not matter, we can use combinations.
We can select 2 men from 4 men in 4C2 ways (6 ways)
Aside: If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Here comes the brute force part.
For every selection of 2 men, determine the number of possible games that can be played.
Let's say the 4 couples are Aa, Bb, Cc, and Dc (where the upper case letter is the husband and the lower case letter is the wife)
So, let's say we picked A and B as the two men.
The possible teams are:
- Ab versus Ba
- Ab versus Bc
- Ab versus Bd
- Ac versus Ba
- Ac versus Bd
- Ad versus Ba
- Ad versus Bc
So, when we picked A and B as the two men, there are 7 possible games to be played.
Since there are 6 different ways to choose the 2 men, the total number of possible games = (6)(7) = [spoiler]42 = D[/spoiler]
Cheers,
Brent
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lunarpower
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Ha!Brent@GMATPrepNow wrote: Hey Ron,
The question says "mixed" tennis, which means each team consists of 1 man and 1 woman.
Oh jee whiz. For crying out loud.
That's just all kinds of wrong for an exam like this.
Unacceptable degree of knowledge -- and class bias, to boot. (... like the infamous "oarsman / regatta" problem that showed up on the SAT a couple of decades ago)
Ron has been teaching various standardized tests for 20 years.
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On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
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Brent@GMATPrepNow
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Agreed.lunarpower wrote: Unacceptable degree of knowledge -- and class bias, to boot. (... like the infamous "oarsman / regatta" problem that showed up on the SAT a couple of decades ago)
It reminded me of a GMAT question (that appeared on the paper-based tests A LONG TIME AGO) that assumed everyone knows that basketball games do not end in ties.
Cheers,
Brent
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GMATGuruNY
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Mixed tennis means each team must consist of a man and woman.Mission2012 wrote:A group of 4 married couples want to play mixed tennis, but each married individual does not want to be on the same team as his/her spouse. How many possible games can be played between two teams?
(A) 12
(B) 21
(C) 36
(D) 42
(E) 46
(Note: the GMAT would never expect a test-taker to understand the meaning of this term.)
Since there are 4 couples, there are a total of 4 men and 4 women.
A BAD team is composed of a MARRIED couple.
First team:
Number of options for the man = 4.
Number of options for the woman = 4.
To combine these options, we multiply:
4*4 = 16.
From these 16 teams, we must subtract the 4 BAD TEAMS (the 4 married couples):
16-4 = 12.
Second team:
Since the first team is composed of two unmarried people -- a man from one married couple and a woman from a different married couple -- only 2 married couples remain.
Number of options for the man = 3. (Since there are 3 men left.)
Number of options for the woman = 3. (Since there are 3 women left.)
To combine these options, we multiply:
3*3 = 9.
From these 9 teams, we must subtract the 2 BAD TEAMS (the 2 remaining married couples):
9-2 = 7.
To combine our options for the first team with our options for the second team, we multiply:
12*7.
Since the ORDER of the teams doesn't matter -- AC-BD is the same game as BD-AC -- we divide by the number of ways the two teams can be ARRANGED (2!):
(12*7)/(2*1) = 42.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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