Last visit was: 23 Apr 2026, 07:09 It is currently 23 Apr 2026, 07:09
Home
Messages
Tests
Schools
Events
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
abhijit_sen
User avatar
Joined: 10 Sep 2007
Last visit: 10 May 2015
Posts: 456
Own Kudos:
958
GMAT 1: 690 Q50 V34
Posts: 456
Kudos: 958
We have 1 pair for each number between 1-6 inclusive. What 05 Jul 2008, 22:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We have 1 pair for each number between 1-6 inclusive. What is the probability that if 4 numbers are selected out of 12 then none of them belong to same pair?

Although all suggestions are welcome, I prefer If someone can solve it using combinatorics.

OA to follow later.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
code19
User avatar
Joined: 23 Jun 2008
Last visit: 05 Feb 2014
Posts: 69
Own Kudos:
69
Given Kudos: 24
Location: Australia
Schools: AGSM '21
GMAT Date: 04-01-2014
Schools: AGSM '21
Posts: 69
Kudos: 69
We have 1 pair for each number between 1-6 inclusive. What Updated on: 07 Jul 2008, 22:51
Originally posted by code19 on 05 Jul 2008, 22:37.
Last edited by code19 on 07 Jul 2008, 22:51, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1st number can be any : 12/12
2nd number : 10/11 (11 left after the first one and 10 of them are eligible)
3rd number : 8/10 (10 left after the first and the second numbers, 8 of them are eligible)
4th number : 6/9 (9 left after 3 selections, 6 of them are eligible)


(12/12) (10/11) (8/10) (6/9) = 16/33
User avatar
abhijit_sen
User avatar
Joined: 10 Sep 2007
Last visit: 10 May 2015
Posts: 456
Own Kudos:
958
GMAT 1: 690 Q50 V34
Posts: 456
Kudos: 958
We have 1 pair for each number between 1-6 inclusive. What 06 Jul 2008, 17:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Any more takers for it.
User avatar
mbawaters
User avatar
Joined: 07 Jan 2008
Last visit: 12 Sep 2010
Posts: 145
Own Kudos:
366
Posts: 145
Kudos: 366
We have 1 pair for each number between 1-6 inclusive. What 15 Jul 2008, 18:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
code19
1st number can be any : 12/12
2nd number : 10/11 (11 left after the first one and 10 of them are eligible)
3rd number : 8/10 (10 left after the first and the second numbers, 8 of them are eligible)
4th number : 6/9 (9 left after 3 selections, 6 of them are eligible)


(12/12) (10/11) (8/10) (6/9) = 16/33
Show more
deto code19!

Whats the OA?
User avatar
alpha_plus_gamma
User avatar
Joined: 14 Aug 2007
Last visit: 17 Jun 2010
Posts: 298
Own Kudos:
641
Concentration: MC, PE, VC
Posts: 298
Kudos: 641
We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
abhijit_sen
We have 1 pair for each number between 1-6 inclusive. What is the probability that if 4 numbers are selected out of 12 then none of them belong to same pair?

Although all suggestions are welcome, I prefer If someone can solve it using combinatorics.

OA to follow later.
Show more

I don't think my answer is right but I tried to solve it by combinatorics.

heres my approach

4 numbers out of 12 = 12C4 = 495
we have six pairs of numbers,

possible of getting 4 numbers such that they belong to pair
= selecting 2 pairs of numbers out of 6 pairs
= 6C1 * 5C1
= 30

probability of getting 4 numbers that belong to pair = 30/495

so,
probabilty of selecting 4 numbers such that none of them belong to same pair
= 1 - probability of getting 4 numbers that belong to pair
= 1 - 30/495
= 465/495
= 31/33

I don't think this is right answer. I have messed up somewhere in between.
User avatar
maratikus
User avatar
Joined: 01 Jan 2008
Last visit: 22 Jul 2010
Posts: 257
Own Kudos:
347
Given Kudos: 1
Posts: 257
Kudos: 347
We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
User avatar
alpha_plus_gamma
User avatar
Joined: 14 Aug 2007
Last visit: 17 Jun 2010
Posts: 298
Own Kudos:
641
Concentration: MC, PE, VC
Posts: 298
Kudos: 641
We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
maratikus
total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
Show more

can you please explain why you choose 4 out of 6 pairs? we have to choose 4 numbers right? so 2 pairs should be selected.
User avatar
abhijit_sen
User avatar
Joined: 10 Sep 2007
Last visit: 10 May 2015
Posts: 456
Own Kudos:
958
GMAT 1: 690 Q50 V34
Posts: 456
Kudos: 958
We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 20:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
maratikus
total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
Show more

Right on point. I was waiting for something like this. You hit the bulls eye.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!