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Six married couples are standing in a room. If 4 people are

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Six married couples are standing in a room. If 4 people are [#permalink]

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New post 16 Aug 2003, 03:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Six married couples are standing in a room.
If 4 people are chosen at random, find the prob. that exactly one married couple is there among the 4 people.

how dyo solve this? id say:
we have to select only one couple so:
1c6*2c10/4c12=6/11... here i would have, however, combinations with
2 couples, so i have to substract these
2 couples=2c6/4c12=1/33

so answer=6/11-1/33=17/33...

but the official answer is 16/33 (in fact if i try to solve it by 1. finding the prob that 0 couples are among the 4 people: 12*10*8*6/4!/(4c12)=16/33; 2. finding the prob that 2 couples are among the 4 people: 2c6/4c12=1/33; and 3. prob 1 couple=1-prob 0 couples - prob 2 couples=1-16/33-1/33=16/33; i get the right anwer) ...

aaargh!!!... what am i missing???... pls help. thx, j

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New post 16 Aug 2003, 14:46
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my approach

FAVORABLE: a four-position group _ _ _ _

1. take any pair and fill two places (6 ways)
2. take another person to fill the third place (5 pairs=10 ways)
3. take the last person (8 ways, for we cannot take a spose of the foregoing person)

TOTAL: 12C4

(6*10*8)/(9*5*11)=32/33

I also made a mistake, but where?

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New post 16 Aug 2003, 15:24
another way i tried to solve this problem was the following:
for the first person i have 12 possibilities
for the second (assuming first two people form couple), only 1
for the third, i have 10 possibilities
for the fourth, i have 8 possibilities
so number of arrangement with one couple=(12*1*10*8)/4!

this yields ((12*1*10*8)/4!)/(12c4/4!)=8/99

aaargh :horror

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New post 16 Aug 2003, 15:35
i think i know where the problem is:
((12*1)/2!)*((10*8)/2!)/12c4=16/33 (right answer)

note the difference with ((12*1*10*8)/4!)/12c4?

however, i cant figure out why is the first approach right...

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Re: probability problem [#permalink]

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New post 19 Aug 2003, 07:43
javropu wrote:
Six married couples are standing in a room.
If 4 people are chosen at random, find the prob. that exactly one married couple is there among the 4 people.

how dyo solve this? id say:
we have to select only one couple so:
1c6*2c10/4c12=6/11... here i would have, however, combinations with
2 couples, so i have to substract these
2 couples=2c6/4c12=1/33

so answer=6/11-1/33=17/33...

but the official answer is 16/33 (in fact if i try to solve it by 1. finding the prob that 0 couples are among the 4 people: 12*10*8*6/4!/(4c12)=16/33; 2. finding the prob that 2 couples are among the 4 people: 2c6/4c12=1/33; and 3. prob 1 couple=1-prob 0 couples - prob 2 couples=1-16/33-1/33=16/33; i get the right anwer) ...

aaargh!!!... what am i missing???... pls help. thx, j


There are 12C4 = 495 ways to pick 4 people.

There are 6 ways to pick one couple, and 10C2 = 45 ways to pick the other two people, 5 of which are couples.

Hence, there are 6*(45-5)=240 ways to pick one couple in 4 people.

Hence the prob is 240/495 = 16/33
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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New post 19 Aug 2003, 09:57
thx akamay... you make it look so easy... its embarrasing... but ive learned a lot, thx... i discussed this problem with another smart guy and he told me this other way:
1c6*2c5*1c2*1c2/4c12
-you pick a couple among the 6, then you pick 2 more couples among the other 5 and then you pick a member of each couple-... cheers, javi

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New post 19 Aug 2003, 12:12
stolyar wrote:
my approach

FAVORABLE: a four-position group _ _ _ _

1. take any pair and fill two places (6 ways)
2. take another person to fill the third place (5 pairs=10 ways)
3. take the last person (8 ways, for we cannot take a spose of the foregoing person)

TOTAL: 12C4

(6*10*8)/(9*5*11)=32/33

I also made a mistake, but where?


You double count in steps 2 and 3.

consider step #1. If you counted it the same way you would say:
1a: pick any person (12 ways)
1b: pick the spouse (1 way)
but there are only 6 pairs! If you pick the husband first then the wife, it is the same combination as if you pick the wife first, then the husband.

Similarly, you pair up each non-couple twice in steps 2 and 3 so you have exactly twice as many combinantions as you need.

Do you get it?
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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New post 19 Aug 2003, 14:03
mmmmm... ive double checked my numbers
1c6*2c5*1c2*1c2/4c12=16/33
this doesnt prove anything, i admit it...
i get your point about the first pair... however, im still missing how does that relate to my approach... ive picked one pair... 6 ways... then i pick 2 pairs out of the 5 left (to make sure i dont pick two people from the same pair)... 10 ways (not 20)... then i pick one different member of each pair... 4 ways (not 6)... so total=6*10*4=240... i just dont see where im double counting... help me see it please, as i cannot see it myself... thk u very much

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New post 19 Aug 2003, 14:41
javropu wrote:
mmmmm... ive double checked my numbers
1c6*2c5*1c2*1c2/4c12=16/33
this doesnt prove anything, i admit it...
i get your point about the first pair... however, im still missing how does that relate to my approach... ive picked one pair... 6 ways... then i pick 2 pairs out of the 5 left (to make sure i dont pick two people from the same pair)... 10 ways (not 20)... then i pick one different member of each pair... 4 ways (not 6)... so total=6*10*4=240... i just dont see where im double counting... help me see it please, as i cannot see it myself... thk u very much


You are fine. I was responding to Stolyar.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: Six married couples are standing in a room. If 4 people are [#permalink]

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Re: Six married couples are standing in a room. If 4 people are   [#permalink] 04 Oct 2017, 01:38
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