It is currently 23 Oct 2017, 14:39

### 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

# 6 married couple are present at a party. If 4 people are

Author Message
Senior Manager
Joined: 06 Jul 2004
Posts: 467

Kudos [?]: 137 [0], given: 0

Location: united states
6 married couple are present at a party. If 4 people are [#permalink]

### Show Tags

07 Aug 2007, 20:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 married couple are present at a party. If 4 people are selected out of these 12, what is the probability that none of these people will be married to each other?

A) 1/33

B) 2/33

C) 1/3

D) 16/33

E) 11/12
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

Kudos [?]: 137 [0], given: 0

Senior Manager
Joined: 06 Jul 2004
Posts: 467

Kudos [?]: 137 [0], given: 0

Location: united states

### Show Tags

07 Aug 2007, 23:43
xsaudx wrote:

I know the answer. It's definitely not A. Would you mind sharing your thought process?
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

Kudos [?]: 137 [0], given: 0

Manager
Joined: 13 May 2007
Posts: 241

Kudos [?]: 21 [0], given: 0

### Show Tags

08 Aug 2007, 00:02
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

Kudos [?]: 21 [0], given: 0

Senior Manager
Joined: 06 Jul 2004
Posts: 467

Kudos [?]: 137 [0], given: 0

Location: united states

### Show Tags

08 Aug 2007, 00:18
empty_spaces wrote:
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

Thanks. This is the right answer.

By the way, I was trying to solve it in the following manner:

number of ways of selecting 4 people out of 12 = 12C4 = 495.

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

This is greater than 495. What is my mistake?

I also tried, finding the number of ways we can pick couples among these 4 people.

First person can be chosen in 12 ways.
Second person can be chosen in only 1 ways.

third person can be chosen in 10 ways.
fourth person can be chosen in 1 way

so the total number of favourable ways = 495 -120=375

P=375/495.

could you please point out my errors?
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

Kudos [?]: 137 [0], given: 0

Manager
Joined: 13 May 2007
Posts: 241

Kudos [?]: 21 [0], given: 0

### Show Tags

08 Aug 2007, 00:54
Let me try :

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

above statement is wrong in this context ....you are mixing permutations and combinations .

if you want solve this way then divide "12*10*8*6" by 12P4 instead of 12C4..and you should have the correct answer.

yet another method :

consider each couple as one and then find : 6C4 for selecting 4 out of 6 couples but now since you only want one out of each of these 4 couples you need to multiply this by 2 for each couple which gives you ...2 * 2 * 2 * 2* 6C4

now your answer is : (2 * 2 * 2 * 2* 6C4 ) / 12 C 4 ...solving that should give you the correct answer

shoonya wrote:
empty_spaces wrote:
probability of choosing the first person : 12/12 = 1

probability of choosing the 2nd person = 10/11 (who is not married to the first one )

probability of choosing 3rnd person = 8/10 (who is not married to either of the first 2 )

probability of choosing 4th person = 6/9 (who is not married to either of the first 3 )

so combining ... 1 * 10/11 * 8/10 * 6/9 = 16/33

Thanks. This is the right answer.

By the way, I was trying to solve it in the following manner:

number of ways of selecting 4 people out of 12 = 12C4 = 495.

now, number of ways of choosing 4 people in such a way that there are no couples

= 12*10*8*6.

This is greater than 495. What is my mistake?

I also tried, finding the number of ways we can pick couples among these 4 people.

First person can be chosen in 12 ways.
Second person can be chosen in only 1 ways.

third person can be chosen in 10 ways.
fourth person can be chosen in 1 way

so the total number of favourable ways = 495 -120=375

P=375/495.

could you please point out my errors?

Kudos [?]: 21 [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 324 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

08 Aug 2007, 02:28
D

P (1st Person) = 12/12 - We can choose anybody
P (2nd Person) = 10/11 - Out of remaining 11 we can choose 10 because we can not choose who is already selected and his/her spouse.
P (3rd Person) = 8/10 - Out of remaining 10 we can choose 8 because we can not choose who are already selected (2 selected) and their 2 spouses.
P (4th Person) = 6/9 - Out of remaining 9 we can choose 6 because we can not choose who are already selected (3 selected) and their 3 spouses.

Total Prob = 1 * 10/11 * 8/10 * 6/9 = 16/33

Kudos [?]: 324 [0], given: 0

08 Aug 2007, 02:28
Display posts from previous: Sort by

# 6 married couple are present at a party. If 4 people are

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.