Last visit was: 12 Dec 2024, 02:35 It is currently 12 Dec 2024, 02:35
Home
GMAT
Messages
Tests
Schools
Reviews
Social
Chat
My Profile
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
Praetorian
User avatar
Joined: 15 Aug 2003
Last visit: 27 Dec 2017
Posts: 2,876
Own Kudos:
1,669
Given Kudos: 781
Posts: 2,876
Kudos: 1,669
Combinations - Groups 11 Sep 2003, 03:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

(N/A)

Question Stats:

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

History

Date
Time
Result
Not Attempted Yet
In how many ways can 4 groups of 2 each be selected from a group of 8 Students?


Is it 8C2 * 6C2 * 4C2 * 2C2 ?

Thanks
Praetorian



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 Problem Solving (PS) 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!
avatar
stolyar
avatar
Joined: 03 Feb 2003
Last visit: 06 May 2014
Posts: 1,012
Own Kudos:
1,751
Posts: 1,012
Kudos: 1,751
Combinations - Groups 12 Sep 2003, 10:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]
User avatar
MartinMag
User avatar
Joined: 22 May 2003
Last visit: 13 Jul 2004
Posts: 187
Own Kudos:
840
Location: Uruguay
Posts: 187
Kudos: 840
Combinations - Groups 12 Sep 2003, 17:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stolyar
I think no. There is the doubling effect: the set of some combinations defines the set of others.

For example, we have a group of 2 people, in how many ways we can divide them? The answer is clear -- there is the only way to do so. However, if we employ combinatorics, we count twice: 2C1=2.

In your case the doubling effect is much more complicated.
I think it is [8C2*6C2*4C2*2C2]/[4!*3!*2!*1!]


Maybe I didn't get precisely what you mean, but the number of ways you can arrange a group of 2 people is 2C2=1

Therefore I agree with Praetorian's solution. (not that I don't have any doubts)
User avatar
AkamaiBrah
User avatar
GMAT Instructor
Joined: 07 Jul 2003
Last visit: 24 Jun 2009
Posts: 392
Own Kudos:
496
Location: New York NY 10024
Concentration: Finance
Schools:Haas, MFE; Anderson, MBA; USC, MSEE
Posts: 392
Kudos: 496
Combinations - Groups 14 Sep 2003, 03:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
praetorian123
In how many ways can 4 groups of 2 each be selected from a group of 8 Students?


Is it 8C2 * 6C2 * 4C2 * 2C2 ?

Thanks
Praetorian


Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.
User avatar
Praetorian
User avatar
Joined: 15 Aug 2003
Last visit: 27 Dec 2017
Posts: 2,876
Own Kudos:
1,669
Given Kudos: 781
Posts: 2,876
Kudos: 1,669
Combinations - Groups 14 Sep 2003, 04:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AkamaiBrah
Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.



That was awesome!
Thanks..My solution gives the same answer.



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 Problem Solving (PS) 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!
Moderators:
Bunuel
Math Expert
97841 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3116 posts