Last visit was: 25 Apr 2026, 20:18 It is currently 25 Apr 2026, 20:18
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
schumacher
Joined: 12 Dec 2004
Last visit: 29 Jul 2006
Posts: 23
Own Kudos:
Posts: 23
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
oxon
Joined: 07 Nov 2004
Last visit: 12 Dec 2004
Posts: 43
Own Kudos:
Location: London
Posts: 43
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
schumacher
Joined: 12 Dec 2004
Last visit: 29 Jul 2006
Posts: 23
Own Kudos:
Posts: 23
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
oxon
Joined: 07 Nov 2004
Last visit: 12 Dec 2004
Posts: 43
Own Kudos:
Location: London
Posts: 43
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I was not asking you to "reveal" the OA.
avatar
rohitprabhu
Joined: 08 Dec 2004
Last visit: 08 Nov 2006
Posts: 1
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!
User avatar
ComplexVision
Joined: 08 Nov 2004
Last visit: 11 May 2005
Posts: 15
Own Kudos:
Location: Montreal
Posts: 15
Kudos: 163
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rohitprabhu
Let me give a try.
Is it 14C3 + 11C3 + 9C2 + 7C2 + 5C2 + 3C2 = 607?

Thanks!



I think this appraoch is correct, but it should be

14C3 + 11C3 + 8C2 + 6C2 + 4C2 + 1 = Whatever
User avatar
Dookie
Joined: 19 May 2004
Last visit: 07 Sep 2006
Posts: 144
Own Kudos:
Posts: 144
Kudos: 35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I got,

(14C3 * 11C3 * 8C2 * 6C2 * 4C2 * 2C2) / 6!
User avatar
twixt
Joined: 31 Aug 2004
Last visit: 03 Jul 2015
Posts: 283
Own Kudos:
Posts: 283
Kudos: 404
Kudos
Add Kudos
Bookmarks
Bookmark this Post
14C3.11C3.8C2.6C2.4C2.2C2/(2!.4!) for me
User avatar
schumacher
Joined: 12 Dec 2004
Last visit: 29 Jul 2006
Posts: 23
Own Kudos:
Posts: 23
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Here's the answer given

14! 1
-------------------- x ---------
3!*3!*2!*2!*2!*2!* 2!*4!

which works out to 3153150

I understand the first part but I don't get the second part.

twixt: can you explain?
User avatar
twixt
Joined: 31 Aug 2004
Last visit: 03 Jul 2015
Posts: 283
Own Kudos:
Posts: 283
Kudos: 404
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shumi,

Groups are not distinct here so you have to divide your comb number by the number of groups !. In this case these groups are not equally sized so you have to consider them as different entities : first you have 2 groups of 3 (so divide by 2!) and then 4 groups of 2 (so divide by 4!)
User avatar
schumacher
Joined: 12 Dec 2004
Last visit: 29 Jul 2006
Posts: 23
Own Kudos:
Posts: 23
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
merci beaucoup monsieur!!



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!