It is currently 19 Oct 2017, 08:16

# STARTING SOON:

Live Chat with Cornell Adcoms in Main Chat Room  |  R1 Interview Invites: MIT Sloan Chat  |  UCLA Anderson Chat  |  Duke Fuqua Chat (EA Decisions)

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Six Married couples are standing in a room. If 4 people are

Author Message
Senior Manager
Joined: 20 Dec 2004
Posts: 252

Kudos [?]: 111 [0], given: 0

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

### Show Tags

08 Jan 2005, 22:15
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:02) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

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 probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.

Kudos [?]: 111 [0], given: 0

Manager
Joined: 06 Sep 2004
Posts: 69

Kudos [?]: [0], given: 0

### Show Tags

09 Jan 2005, 06:57
(a) 15 (b) 240 (c) 480

(a) 6C2
(b) 16 * 6C4
(c) 8 * 6C1 * 5C3

Kudos [?]: [0], given: 0

Manager
Joined: 31 Aug 2004
Posts: 161

Kudos [?]: 15 [0], given: 0

### Show Tags

10 Jan 2005, 00:04
I have trouble with this question, help anyone?

Kudos [?]: 15 [0], given: 0

Manager
Joined: 19 Nov 2004
Posts: 59

Kudos [?]: 7 [0], given: 0

### Show Tags

11 Jan 2005, 04:32
a) 6C2/12C4=1/33
b) (12C1x10C1x8C1x6C1)/(12C4x4!)=16/33
c) 1-a)-b)=1-17/33=16/33

OR (6C1x10C1x8C1)/(12C4x2!)=16/33
Sure there is easier way but this is the first solution came into my head
If you need more explanation, I'll give them later

Kudos [?]: 7 [0], given: 0

Manager
Joined: 31 Aug 2004
Posts: 161

Kudos [?]: 15 [0], given: 0

### Show Tags

11 Jan 2005, 10:10
lovely_baby wrote:
a) 6C2/12C4=1/33
b) (12C1x10C1x8C1x6C1)/(12C4x4!)=16/33
c) 1-a)-b)=1-17/33=16/33

OR (6C1x10C1x8C1)/(12C4x2!)=16/33
Sure there is easier way but this is the first solution came into my head
If you need more explanation, I'll give them later

So for 1), we treat each couple as one unit, and we need to pick 2 out of this six units. We get 6C2, and 12C4 means the number of combinations we can get from 12 people. Ok, this one I get it.

For 2), what is the logic to use 12C1 x 10C1 x 8C1 x 6C1 and 12C4 x 4!?

Kudos [?]: 15 [0], given: 0

Manager
Joined: 19 Nov 2004
Posts: 59

Kudos [?]: 7 [0], given: 0

### Show Tags

12 Jan 2005, 00:06
b) no married couple among 4 people
first person you can choose in 12C1 ways (any of 12 people)
second person 10C1 (any of left but not the spouse of the first chosen 11-1=10)
third the same logic (neither the spouse of the first nor the second's) 8C1
fourth 6C1
But to exclude the same fourths with different order we divide it by 4!
To get the probability we should divide it by the number of all possible fourths 12C4

tried to do my best. Did it help?

Kudos [?]: 7 [0], given: 0

Senior Manager
Joined: 22 Jun 2004
Posts: 391

Kudos [?]: 9 [0], given: 0

Location: Bangalore, India

### Show Tags

12 Jan 2005, 06:38
It is tough for me too. I agree with the answers given by lovely_baby.

(a) 6C2/12C4 = 1/33
(b)The number of ways the chosen 4 can be unmarried = 12C1 * 10 C1 * 8C1 * 6C1 = 12*10*8*6

Because the above number is a permutation (order is important), we need to convert it into combination (order is not important as we are choosing) i.e., The actual ways without regard for order = (12*10*8*6)/4!
= 5*8*6
The required probability = (5*8*6)/12C4 = 16/33

(c) The probability that exactly one is married couple = 1-(a)-(b) = 16/33

Can anybody solve (c) in the way we have solved (a) & (b)? Somehow, I am in the loop and unable to come out of it.

neelesh wrote:
This question is probably a simple one for most people on this forum here, but wanted to see what approach to use for such questions. Somehow my appeoach is not matching with the OA.

Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Kudos [?]: 9 [0], given: 0

Manager
Joined: 19 Nov 2004
Posts: 59

Kudos [?]: 7 [0], given: 0

### Show Tags

12 Jan 2005, 08:11
Especially for mallelac!!!
From my previous message
c) (6C1x10C1x8C1)/(12C4x2!)=16/33
6C1 one of 6 couples
10C1 one of left 10 people
8C1 one more of left 8
divide by 2! because doesn't matter the order

Kudos [?]: 7 [0], given: 0

Senior Manager
Joined: 22 Jun 2004
Posts: 391

Kudos [?]: 9 [0], given: 0

Location: Bangalore, India

### Show Tags

12 Jan 2005, 08:18
Thanks for the answer. But lovely_baby,

We need to divide by 3! because the number of ways with 3 items (1 couple and 2 members who are not a couple) are calculated here and order is not important here. infact I too got the answer with 2! division but could not justify it.

lovely_baby wrote:
Especially for mallelac!!!
From my previous message
c) (6C1x10C1x8C1)/(12C4x2!)=16/33
6C1 one of 6 couples
10C1 one of left 10 people
8C1 one more of left 8
divide by 2! because doesn't matter the order

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Kudos [?]: 9 [0], given: 0

Manager
Joined: 19 Nov 2004
Posts: 59

Kudos [?]: 7 [0], given: 0

### Show Tags

12 Jan 2005, 08:24

I don't know how to explain better in English, but in this case the couple doesn't matter. You just choose one and forget about it . But while choosing the singles you twice them...mmm... is it a bit more clear?

Kudos [?]: 7 [0], given: 0

Intern
Joined: 09 Nov 2004
Posts: 7

Kudos [?]: [0], given: 0

Location: Vancouver

### Show Tags

26 Feb 2005, 13:10
I see it like this:

c) exactly one married couple is among the 4.

A A, B B, C C, .., E E
twelve people; six couples

Favorable outcome (example):
A A B C

6C1 ways to select one couple (A A)

10C2 ways to select the remining two people (B C) - but this includes selecting another couple (like B B, or C C); so subtract 5C1

Favorable combinations = 6C1*(10C2 - 5C1) = 240
Total combinations = 12C4 = 495
P(AABC) = 240/490 = 16/33

Kudos [?]: [0], given: 0

Manager
Joined: 27 Jan 2005
Posts: 100

Kudos [?]: 1 [0], given: 0

Location: San Jose,USA- India

### Show Tags

27 Feb 2005, 01:50
I dont understand 8*6C1*5C3? Can you explain how it came ?

Kudos [?]: 1 [0], given: 0

Senior Manager
Joined: 22 Jun 2004
Posts: 391

Kudos [?]: 9 [0], given: 0

Location: Bangalore, India

### Show Tags

27 Feb 2005, 03:03
Excellent!! I got the stuff very clearly. I was getting the answer from set theory but not in a direct way.

Thanks d-dogg!!

d-dogg wrote:
I see it like this:

c) exactly one married couple is among the 4.

A A, B B, C C, .., E E
twelve people; six couples

Favorable outcome (example):
A A B C

6C1 ways to select one couple (A A)

10C2 ways to select the remining two people (B C) - but this includes selecting another couple (like B B, or C C); so subtract 5C1

Favorable combinations = 6C1*(10C2 - 5C1) = 240
Total combinations = 12C4 = 495
P(AABC) = 240/490 = 16/33

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Kudos [?]: 9 [0], given: 0

VP
Joined: 30 Sep 2004
Posts: 1480

Kudos [?]: 411 [0], given: 0

Location: Germany

### Show Tags

28 Feb 2005, 05:39
HongHu wrote:
ssumitsh wrote:
(c) 8 * 6C1 * 5C3

What's wrong with this?

can you plz explain the second and the third answer ? what would be the best concept to solve them ? thx

Kudos [?]: 411 [0], given: 0

Senior Manager
Joined: 22 Jun 2004
Posts: 391

Kudos [?]: 9 [0], given: 0

Location: Bangalore, India

### Show Tags

28 Feb 2005, 06:18
Christoph,

It is tough for me too. I agree with the answers given by lovely_baby.

(a) 6C2/12C4 = 1/33
(b) While choosing 4, no couple shoule be present. Let us pick 1 out of 12. In the remaining 12, the first person's spouse is present and thus has to be skiiped because no couple should be present in 4. So, we can choose 2nd person out of 10. This 2nd person's spouse is present in the remaining 9. So, the 3rd person has to be from the remaining 8. Similarly, 4th person has to be from the remaining 6.

The total number of ways the chosen 4 can form no couples = 12*10*8*6
Here, all the number of ways are accounted i.e. ABCD, ABDC, ADCB etc. But, when we pick 4 persons at random, they can be in any order. All of ABCD, ABDC, ADCB etc are one and the same as for as the group is concerned. Thus, we need to divide the (12*10*8*6) with 4!

In other words,
Because the above number is a permutation (order is important), we need to convert it into combination (order is not important as we are choosing) i.e., The actual ways without regard for order = (12*10*8*6)/4!
= 5*8*6
The required probability = (5*8*6)/12C4 = 16/33

(c) If you observe care fully, 1 = P(a) + P(b) + P(c) from the set theory because Total probability (i.e., 1) = Probability that two married couples are chosen + Probability that one married couple is chosen + Probability that NO married couple is chosen

The probability that exactly one is married couple = 1-P(a)-P(b) = 16/33

The above method is easy by observation. However, if P(c) is asked straight, then you may work out as follows.

Let P(c) = Probability that exactly one is married

The number of ways of choosing 4 such that only 1 is married couple = 6C1 * 10C2 - 6C1*5C1 = 6*45 - 6 * 5 = 240

This is very well explained d-dogg

"6C1 ways to select one couple (A A)

10C2 ways to select the remining two people (B C) - but this includes selecting another couple (like B B, or C C); so subtract 5C1

Favorable combinations = 6C1*(10C2 - 5C1) = 240 "

Total space = 12C4

P(c) = 240/12C4 = 16/33

The second method is easier if you understand the logic in it. It is faster too. If you are uncomfortable with it, go for the first method where you should be good at breaking up the unknown into known problems.

neelesh wrote:
This question is probably a simple one for most people on this forum here, but wanted to see what approach to use for such questions. Somehow my appeoach is not matching with the OA.

Six Married couples are standing in a room. If 4 people are chosen at random, find the probability p for the following scenarios

a) 2 married couples are chosen
b) no married couple is among the 4
c) exactly one married couple is among the 4.

christoph wrote:
HongHu wrote:
ssumitsh wrote:
(c) 8 * 6C1 * 5C3

What's wrong with this?

can you plz explain the second and the third answer ? what would be the best concept to solve them ? thx

_________________

Awaiting response,

Thnx & Rgds,
Chandra

Kudos [?]: 9 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

28 Feb 2005, 08:36
This is a very good question. Permutation and combination questions often have multiple solutions that could reach the same answer. Allow yourself to think more than just gets the answer will give yourself an opportunity to really understand the concept and master the skills to solve this kind of questions. That is why I would like to nudge people to think more.

Love_baby has a good approach and he had explained it clearly. Ssumitsh also has a good approach. He only answered the number of outcomes, but it can be easily converted to probabilities. However, his answer for (c) is different from the right answer. Why?

700Plus wrote:
I dont understand 8*6C1*5C3? Can you explain how it came ?

To understand his logic, I would explain his solution to (b) and see if anybody can understand his logic for (c) and found why he gave a wrong answer.

ssumitsh wrote:
(b) 16 * 6C4

(b) asks for probability that non of the four people are couples. So you first pick four couples out of the six couples. C(6,4) Then you get one person from each of the four picked couples. For each couple you have two choices. (C(2,1)=2)
Therefore the combined outcome is: C(6,4)*2*2*2*2=240

Now we know that (c) is wrong. But his line of thought is right. So what did he do wrong?

Kudos [?]: 376 [0], given: 0

VP
Joined: 30 Sep 2004
Posts: 1480

Kudos [?]: 411 [0], given: 0

Location: Germany

### Show Tags

28 Feb 2005, 10:56
thx honguhu and mallelac !

how do we know the number of permutations respectively 4! in B) and 2! in A) ?

Kudos [?]: 411 [0], given: 0

Manager
Joined: 24 Jan 2005
Posts: 217

Kudos [?]: 13 [0], given: 0

Location: Boston

### Show Tags

28 Feb 2005, 15:43
(a) Choose the first person in 12 ways, the next person chosen in 1 way (his partner), the 3rd person in 10 ways and then the last person (his partner) in one way so 10*12 = 120 ways

(b) 12 ways * 10 ways * 8 ways * 6 ways (skipping one partner for each)
= 5760

(c) 12 ways * 1 * 10 ways * 8 ways = 960

Not confident with my approach though

Kudos [?]: 13 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

28 Feb 2005, 15:46
Don't forget the difference bt combinations and permutations, anirban16.

Kudos [?]: 376 [0], given: 0

28 Feb 2005, 15:46
Display posts from previous: Sort by