Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 03 Feb 2010
Posts: 68

Seven men and seven women have to sit around a circular [#permalink]
Show Tags
07 Apr 2010, 16:10
1
This post received KUDOS
7
This post was BOOKMARKED
Question Stats:
60% (01:31) correct
40% (02:39) wrong based on 27 sessions
HideShow timer Statistics
Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?



Intern
Joined: 07 Apr 2010
Posts: 25

Re: Circular probability question? [#permalink]
Show Tags
07 Apr 2010, 21:06
3
This post received KUDOS
7 men can sit around a circular table in (71)! ways = 6! [Logic: no: of asymmetric circular permutations of n objects is (n1)!.] Next, all you need to do is seat the women in the vacant 7 slots (b/w the men) which can be done in 7! ways so, my ans is (6! x 7!) ways
_________________
GMAT, here i come... impressed?...how bout encouraging me with a kudos
cheers just another idiot!



Intern
Joined: 18 Mar 2010
Posts: 1

Re: Circular probability question? [#permalink]
Show Tags
07 Apr 2010, 22:10
Just one question here.........since you will be seating the women also round the circular table, then why is that the logic of no. of asymmetric circular permutations of n objects does not apply here...?



Intern
Joined: 07 Apr 2010
Posts: 25

Re: Circular probability question? [#permalink]
Show Tags
08 Apr 2010, 07:27
since we have already arranged the 7 men in a circular fashion, the question, thereafter, ceases to be based on circular permutation. [Tip : when you consider such circular scenarios, imagine a "passingtheparcel" game in either clockwise or anticlockwise direction. the relative order in which the parcels are passed should be unique among all permutations] Still not convinced?...have a look at my example in the attachment.
_________________
GMAT, here i come... impressed?...how bout encouraging me with a kudos
cheers just another idiot!



Manager
Joined: 10 Aug 2009
Posts: 123

Re: Circular probability question? [#permalink]
Show Tags
08 Apr 2010, 12:37
I would have gone with 7!*7!... can someone please explain why its 6! and not 7! for the men?



Manager
Joined: 07 Jan 2010
Posts: 240

Re: Circular probability question? [#permalink]
Show Tags
09 Apr 2010, 23:05



Manager
Joined: 27 Jul 2010
Posts: 194
Location: Prague
Schools: University of Economics Prague

Re: Circular probability question? [#permalink]
Show Tags
20 Jan 2011, 09:47
idiot wrote: 7 men can sit around a circular table in (71)! ways = 6! [Logic: no: of asymmetric circular permutations of n objects is (n1)!.] Next, all you need to do is seat the women in the vacant 7 slots (b/w the men) which can be done in 7! ways so, my ans is (6! x 7!) ways Are you sure about that?? Why are women not considerad also circular???
_________________
You want somethin', go get it. Period!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7446
Location: Pune, India

Re: Circular probability question? [#permalink]
Show Tags
20 Jan 2011, 21:31
4
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
craky wrote: idiot wrote: 7 men can sit around a circular table in (71)! ways = 6! [Logic: no: of asymmetric circular permutations of n objects is (n1)!.] Next, all you need to do is seat the women in the vacant 7 slots (b/w the men) which can be done in 7! ways so, my ans is (6! x 7!) ways Are you sure about that?? Why are women not considerad also circular??? Yes, the solution given above is correct. Think of it this way: There are 7 men: Mr. A, Mr. B ..... and 7 women: Ms. A, Ms. B .... 14 seats around a circular table. You seat the 7 women such that no two of them are together so they occupy 7 nonadjacent places in 6! ways. For the first woman who sits, each seat is identical. Once she sits, each seat becomes unique and when the next woman sits, she sits in a position relative to the first woman (e.g. 1 seat away on left, 3 seats away on right etc) The 7 men have 7 unique seats to occupy. Each of the 7 seats are unique because they have a fixed relative position (e.g. between Ms. A and Ms. B or between Ms. C and Ms. B etc...). So the men can sit in 7! ways. Total 6!*7! ways.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Sep 2009
Posts: 39723

Re: Circular probability question? [#permalink]
Show Tags
21 Jan 2011, 12:58
3
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
craky wrote: idiot wrote: 7 men can sit around a circular table in (71)! ways = 6! [Logic: no: of asymmetric circular permutations of n objects is (n1)!.] Next, all you need to do is seat the women in the vacant 7 slots (b/w the men) which can be done in 7! ways so, my ans is (6! x 7!) ways Are you sure about that?? Why are women not considerad also circular??? The number of arrangements of n distinct objects in a row is given by n!. The number of arrangements of n distinct objects in a circle is given by (n1)!. The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: n!/n=(n1)! Now, 7 men in a circle can be arranged in (71)! ways and if we place 7 women in empty slots between them then no two women will be together. The # of arrangement of these 7 women will be 7! and not 6! because if we shift them by one position we'll get different arrangement because of the neighboring men. So the answer is indeed 6!*7!. Similar questions: anothertrickycircularpermutationproblem106928.htmlcircularpermutationproblem106919.htmlcirculartable106485.htmlcombinationsproblem104101.htmlarrangementsaroundthetable102184.htmlarrangementinacircle98185.htmlpleasehelpwiththeseatingarrangementproblems94915.htmlHope it helps.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16007

Re: Seven men and seven women have to sit around a circular [#permalink]
Show Tags
21 Sep 2013, 11:51
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16007

Re: Seven men and seven women have to sit around a circular [#permalink]
Show Tags
13 Dec 2014, 23:08
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16007

Re: Seven men and seven women have to sit around a circular [#permalink]
Show Tags
31 May 2016, 06:18
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Manager
Joined: 28 Apr 2016
Posts: 56

Re: Seven men and seven women have to sit around a circular [#permalink]
Show Tags
12 Jul 2016, 07:00
But won't the women also be sitting in a circular manner, hence (71)!? so shouldn't the answer be 6! x 6! ? Bunuel wrote: craky wrote: idiot wrote: 7 men can sit around a circular table in (71)! ways = 6! [Logic: no: of asymmetric circular permutations of n objects is (n1)!.] Next, all you need to do is seat the women in the vacant 7 slots (b/w the men) which can be done in 7! ways so, my ans is (6! x 7!) ways Are you sure about that?? Why are women not considerad also circular??? The number of arrangements of n distinct objects in a row is given by n!. The number of arrangements of n distinct objects in a circle is given by (n1)!. The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: n!/n=(n1)! Now, 7 men in a circle can be arranged in (71)! ways and if we place 7 women in empty slots between them then no two women will be together. The # of arrangement of these 7 women will be 7! and not 6! because if we shift them by one position we'll get different arrangement because of the neighboring men. So the answer is indeed 6!*7!. Similar questions: anothertrickycircularpermutationproblem106928.htmlcircularpermutationproblem106919.htmlcirculartable106485.htmlcombinationsproblem104101.htmlarrangementsaroundthetable102184.htmlarrangementinacircle98185.htmlpleasehelpwiththeseatingarrangementproblems94915.htmlHope it helps.




Re: Seven men and seven women have to sit around a circular
[#permalink]
12 Jul 2016, 07:00







