Last visit was: 22 Apr 2026, 12:47 It is currently 22 Apr 2026, 12:47
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
enfinity
Joined: 12 May 2009
Last visit: 12 Oct 2014
Posts: 41
Own Kudos:
Given Kudos: 18
Posts: 41
Kudos: 241
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
coelholds
Joined: 03 Jul 2009
Last visit: 09 Oct 2013
Posts: 81
Own Kudos:
604
 [1]
Given Kudos: 13
Location: Brazil
Posts: 81
Kudos: 604
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
enfinity
Joined: 12 May 2009
Last visit: 12 Oct 2014
Posts: 41
Own Kudos:
Given Kudos: 18
Posts: 41
Kudos: 241
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
coelholds
Joined: 03 Jul 2009
Last visit: 09 Oct 2013
Posts: 81
Own Kudos:
604
 [1]
Given Kudos: 13
Location: Brazil
Posts: 81
Kudos: 604
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1) Yes, you are right

2) Yes you are right in the first part. The 8 is because the 2 people precluded in 1)

You still must divide by two by two, because when you calculate using C, it already gives you (A,E) but not (E,A). However, you are multiplying and in doing this you are not using the C formula any more, and thus you are counting pairA x pairB twice.

Imagine that you have only 3 PAIRS (not players), X, Y, Z. How many games it is possible?

If you do 3*2 = 6 you are wrong, because you are counting some games twice:
X x Y
X x Z
Y x X
Y x Z
Z x X
Z x Y

As you can see, the bold ones are repeated. That is why you must divide by two. You are not using the formula any more. The formula of Combination is just use in the first part of the problem.

Now is it clear?

If not, you can just ask, ok? Sometimes it is hard to explain only through text.... :)
User avatar
enfinity
Joined: 12 May 2009
Last visit: 12 Oct 2014
Posts: 41
Own Kudos:
Given Kudos: 18
Posts: 41
Kudos: 241
Kudos
Add Kudos
Bookmarks
Bookmark this Post
coelholds! That's a very good explanation! Thanks a lot!

Life is good :)
User avatar
irajeevsingh
Joined: 27 Jun 2008
Last visit: 05 Aug 2009
Posts: 72
Own Kudos:
Given Kudos: 11
Posts: 72
Kudos: 42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Yeah AB vs CD is equal to CD vs AB. Divide by 2 again.
User avatar
lgon
Joined: 14 Nov 2008
Last visit: 25 Jun 2012
Posts: 108
Own Kudos:
Given Kudos: 3
Concentration: Entrepreneurship
Schools:Stanford...Wait, I will come!!!
Posts: 108
Kudos: 658
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks a lot.
Can you suggest any reference material to improve the permutation and combination?
Thank in advance
coelholds
1) Yes, you are right

2) Yes you are right in the first part. The 8 is because the 2 people precluded in 1)

You still must divide by two by two, because when you calculate using C, it already gives you (A,E) but not (E,A). However, you are multiplying and in doing this you are not using the C formula any more, and thus you are counting pairA x pairB twice.

Imagine that you have only 3 PAIRS (not players), X, Y, Z. How many games it is possible?

If you do 3*2 = 6 you are wrong, because you are counting some games twice:
X x Y
X x Z
Y x X
Y x Z
Z x X
Z x Y

As you can see, the bold ones are repeated. That is why you must divide by two. You are not using the formula any more. The formula of Combination is just use in the first part of the problem.

Now is it clear?

If not, you can just ask, ok? Sometimes it is hard to explain only through text.... :)
User avatar
coelholds
Joined: 03 Jul 2009
Last visit: 09 Oct 2013
Posts: 81
Own Kudos:
Given Kudos: 13
Location: Brazil
Posts: 81
Kudos: 604
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sorry lgon, all the math material that I have is in Portuguese.
User avatar
enfinity
Joined: 12 May 2009
Last visit: 12 Oct 2014
Posts: 41
Own Kudos:
Given Kudos: 18
Posts: 41
Kudos: 241
Kudos
Add Kudos
Bookmarks
Bookmark this Post
https://www.4GMAT.com offers a combinations and probablility e-book at $6.99. Does anyone have some experience with 4GMAT ebooks? I feel that they're providing very hard questions but the written English is pretty bad sometimes (which probably confuses ppl when they try to understand the question. Plus, GMAT questions are just worded differently)



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!
Moderator:
Math Expert
109753 posts