Tennis

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Mission2012: Master | Next Rank: 500 Posts; Posts: 184; Joined: Sun Aug 19, 2012 10:04 pmFollowed by: 2 members; Thanked: 10 times

by Mission2012 » Fri Aug 16, 2013 10:23 pm

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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Source: Beat The GMAT — Problem Solving | Fri Aug 16, 2013 10:23 pm

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Sat Aug 17, 2013 5:39 am

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...

Ron has been teaching various standardized tests for 20 years.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Sat Aug 17, 2013 5:40 am

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?

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Aug 17, 2013 5:48 am

lunarpower wrote:

Hey Ron,

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: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Aug 17, 2013 6:01 am

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

Here's one approach:

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

Brent Hanneson - Creator of GMATPrepNow.com

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Sat Aug 17, 2013 6:50 am

Brent@GMATPrepNow wrote: Hey Ron,

The question says "mixed" tennis, which means each team consists of 1 man and 1 woman.

Ha!
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)

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Aug 17, 2013 6:55 am

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)

Agreed.
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

Brent Hanneson - Creator of GMATPrepNow.com

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by GMATGuruNY » Sat Aug 17, 2013 7:13 am

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

Mixed tennis means each team must consist of a man and woman.
(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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