Last visit was: 21 Apr 2026, 19:06 It is currently 21 Apr 2026, 19:06
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
Caas
User avatar
Joined: 04 Mar 2007
Last visit: 13 Aug 2007
Posts: 180
Own Kudos:
864
Posts: 180
Kudos: 864
5 people, including A and B, are to be seated in a row. How 05 May 2007, 12:32
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
5 people, including A and B, are to be seated in a row. How many different ways can they be seated subject to a constraint that A and B can not sit next to each other?

A) 120
B) 96
C) 72
D) 48



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
kekc
User avatar
Joined: 01 May 2007
Last visit: 17 Dec 2009
Posts: 8
Own Kudos:
2
Schools:INSEAD, Wharton, Chicago
Posts: 8
Kudos: 2
5 people, including A and B, are to be seated in a row. How Updated on: 05 May 2007, 14:35
Originally posted by kekc on 05 May 2007, 14:03.
Last edited by kekc on 05 May 2007, 14:35, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
answer = all permutations - constraint
all permutations = 5!
error here: constraint = 8 (ABxxx, BAxxx, xABxx, xBAxx, xxABx, xxBAx, xxxAB, xxxBA)
answer = 5! - 8 = 5*4*3*2 - 8 = 120 - 8 = 112
User avatar
Caas
User avatar
Joined: 04 Mar 2007
Last visit: 13 Aug 2007
Posts: 180
Own Kudos:
864
Posts: 180
Kudos: 864
5 people, including A and B, are to be seated in a row. How 05 May 2007, 14:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kekc
answer = all permutations - constraint
all permutations = 5!
constraint = 8 (ABxxx, BAxxx, xABxx, xBAxx, xxABx, xxBAx, xxxAB, xxxBA)
answer = 5! - 8 = 5*4*3*2 - 8 = 120 - 8 = 112
Show more


constrain is much bigger
you can not assume that all other people are x
they are x,y and z
So only for your first option there will be:
ABxyz,ABxzy,ABzyx etc etc.... :)
User avatar
vikramjit_01
User avatar
Joined: 02 Jan 2007
Last visit: 22 Aug 2009
Posts: 162
Own Kudos:
41
Posts: 162
Kudos: 41
5 people, including A and B, are to be seated in a row. How 05 May 2007, 14:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
E!
Can't prove me wrong there :lol:
but I'll also go with C

A&B never together = Total - A&B always together
Total =5! =120
A&B always together = 4!2! =48
(for A&B always together, treat A&B as one unit so you have total 4 units and hence the 4!, but A&B can switch places amongst themselves and so the 2!)

120 - 48 = 72
and that's both E and C :wink:
User avatar
Caas
User avatar
Joined: 04 Mar 2007
Last visit: 13 Aug 2007
Posts: 180
Own Kudos:
864
Posts: 180
Kudos: 864
5 people, including A and B, are to be seated in a row. How 05 May 2007, 14:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
yeah.... I forgot to post E :roll:
but it doesn't matter :-D
User avatar
kekc
User avatar
Joined: 01 May 2007
Last visit: 17 Dec 2009
Posts: 8
Own Kudos:
2
Schools:INSEAD, Wharton, Chicago
Posts: 8
Kudos: 2
5 people, including A and B, are to be seated in a row. How 05 May 2007, 14:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Caas, you are right, in my solution constraint should be multiplied by 6.
So, 5! - 6*8 = 120 - 48 = 72.
User avatar
[email protected]
User avatar
Joined: 14 Jan 2007
Last visit: 04 Feb 2009
Posts: 312
Own Kudos:
955
Posts: 312
Kudos: 955
5 people, including A and B, are to be seated in a row. How 05 May 2007, 14:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
It should be 'C'.

Total number of ways to seat 5 people in a row = 5! = 120
number of ways A and B will seat with each other = 4!2! = 48
So number of ways to seat 5 people when A and B don't sit next to each other = 120-48 =72



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