Last visit was: 21 Apr 2026, 20:51 It is currently 21 Apr 2026, 20:51
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
boksana
User avatar
Joined: 01 May 2004
Last visit: 31 Oct 2006
Posts: 232
Own Kudos:
382
Location: USA
Posts: 232
Kudos: 382
There are 8 people who want to divide themselves to play 04 Jul 2004, 14:26
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
There are 8 people who want to divide themselves to play double tennis into 4 teams. How many combinations are possible?



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!
User avatar
boksana
User avatar
Joined: 01 May 2004
Last visit: 31 Oct 2006
Posts: 232
Own Kudos:
382
Location: USA
Posts: 232
Kudos: 382
There are 8 people who want to divide themselves to play 04 Jul 2004, 14:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
i don't think so :roll:
User avatar
Paul
User avatar
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,707
Own Kudos:
1,651
Posts: 2,707
Kudos: 1,651
There are 8 people who want to divide themselves to play 04 Jul 2004, 14:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First team: 8C2
Second team: 6C2
Third team: 4C2
Fourth team: 2C2
total outcome: 8C2*6C2*4C2*2C2 = 2520
User avatar
boksana
User avatar
Joined: 01 May 2004
Last visit: 31 Oct 2006
Posts: 232
Own Kudos:
382
Location: USA
Posts: 232
Kudos: 382
There are 8 people who want to divide themselves to play 04 Jul 2004, 14:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Nope Paul! Try to solve for 4 people and you'll see your mistake
User avatar
boksana
User avatar
Joined: 01 May 2004
Last visit: 31 Oct 2006
Posts: 232
Own Kudos:
382
Location: USA
Posts: 232
Kudos: 382
There are 8 people who want to divide themselves to play 04 Jul 2004, 14:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Are you kidding? :)
User avatar
Paul
User avatar
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,707
Own Kudos:
1,651
Posts: 2,707
Kudos: 1,651
There are 8 people who want to divide themselves to play 04 Jul 2004, 15:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
:stupid I think you have to divide by 4! because each team could be chosen 1st, 2nd, 3rd or 4th irrespective of order. Hence 2520/4! = 105
User avatar
Paul
User avatar
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,707
Own Kudos:
1,651
Posts: 2,707
Kudos: 1,651
There are 8 people who want to divide themselves to play 04 Jul 2004, 15:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
:stupid2 :shocked :hammer
User avatar
gmatblast
User avatar
Joined: 11 Nov 2003
Last visit: 27 Dec 2004
Posts: 190
Own Kudos:
68
Location: Illinois
Posts: 190
Kudos: 68
There are 8 people who want to divide themselves to play 10 Aug 2004, 11:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Would it be appropriate to say that if the question had been for the number of different possible teams, then Paul's answer would have been correct?

Here the question asks for how many different combinations? I believe that if the 4 teams are formed as (1,2)(3,4),(5,6) and (7,8), then this whole configuration will constitute one combination.

But as for the number of teams, it is 4 teams

Am I correct?



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:
Bunuel
Math Expert
109728 posts