Seven men and seven women have to sit around a circular : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 12:49

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Seven men and seven women have to sit around a circular

Author Message
TAGS:

### Hide Tags

Intern
Joined: 08 Nov 2004
Posts: 46
Location: Montreal
Followers: 0

Kudos [?]: 23 [2] , given: 0

Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

10 Nov 2004, 09:42
2
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

50% (01:45) correct 50% (01:01) wrong based on 647 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 ways can that be done?

A. 5!*6!
B. 6!*6!
C. 5!*7!
D. 6!*7!
E. 7!*7!
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Feb 2014, 03:50, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [11] , given: 10583

Re: Seven men and seven women [#permalink]

### Show Tags

19 Feb 2012, 04:47
11
KUDOS
Expert's post
16
This post was
BOOKMARKED
ComplexVision wrote:
Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many ways can that be done?

A. 24
B. 6
C. 4
D. 12
E. 3

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 (n-1)!.

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=(n-1)!

Now, 7 men in a circle can be arranged in (7-1)! 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:
another-tricky-circular-permutation-problem-106928.html
circular-permutation-problem-106919.html
circular-table-106485.html
combinations-problem-104101.html
arrangements-around-the-table-102184.html
arrangement-in-a-circle-98185.html

Hope it helps.
_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 594 [1] , given: 355

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

06 Feb 2014, 08:22
1
KUDOS
bumpbot wrote:
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.

Hi Bumpot, I think what you could do for now is change the answer choices for this question so that we can answer it accordingly since the answer choices given don't make much sense

Cheers
J
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [1] , given: 10583

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

07 Feb 2014, 03:50
1
KUDOS
Expert's post
jlgdr wrote:
bumpbot wrote:
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.

Hi Bumpot, I think what you could do for now is change the answer choices for this question so that we can answer it accordingly since the answer choices given don't make much sense

Cheers
J

Thank you for the suggestion. Done.
_________________
Director
Joined: 16 Jun 2004
Posts: 893
Followers: 3

Kudos [?]: 59 [0], given: 0

### Show Tags

10 Nov 2004, 10:27
1
This post was
BOOKMARKED
I am unsure of this - after seeing the given answer choices. I dont get any of the answer choices and also way out from any of the ans choices.

Here is my understanding. I will be glad if someone can let me know the logical error I am committing here.

Putting one Man in a fixed position, the remaining men can be arranged in 6! ways.

For each arrangement, there are 7 positions for the women and they can be arranged in 7! ways.

Meaning the total arrangement is 7!*6! = 3628800 ways - only few hundred thousand ways more than the answer choices. Where am I going wrong?
Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

10 Nov 2004, 11:10
Thats a tough one.

We have to assume clock and anticlockwise arrangements are the same.
By saying that no two women must seat next to one another means that men and women must alternate in their sitting arrangements.

Let us take the example of 3 men [m1,m2,m3] and 3 women [w1,w2,w3]. That gives us (3-1)= 2 arrangements
[w1 m1 w2 m2 w3 m3 w1]
[w1 m2 w2 m1 w3 m3 w1]

Similarly for 7 men and 7 women we must have (7-1) = 6 combinations.

So i'd go for answer B. Anyway, that is how I would do it should this be an actual GMAT question.
Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 79 [0], given: 0

### Show Tags

10 Nov 2004, 12:33
venksune wrote:
I am unsure of this - after seeing the given answer choices. I dont get any of the answer choices and also way out from any of the ans choices.

Here is my understanding. I will be glad if someone can let me know the logical error I am committing here.

Putting one Man in a fixed position, the remaining men can be arranged in 6! ways.

For each arrangement, there are 7 positions for the women and they can be arranged in 7! ways.

Meaning the total arrangement is 7!*6! = 3628800 ways - only few hundred thousand ways more than the answer choices. Where am I going wrong?

I checked this one venskune, and you are right. This is how even I did it. I have this book which is used for entrance exams ( those competitive exams). and this is how it ihas been done. The OAs are wrong...
_________________

Let's get it right!!!!

Director
Joined: 07 Nov 2004
Posts: 689
Followers: 6

Kudos [?]: 142 [0], given: 0

### Show Tags

10 Nov 2004, 12:41
oxon wrote:
Let us take the example of 3 men [m1,m2,m3] and 3 women [w1,w2,w3]. That gives us (3-1)= 2 arrangements
[w1 m1 w2 m2 w3 m3 w1]
[w1 m2 w2 m1 w3 m3 w1]

Oxon for 3 men and 3 women example we actually have 12 ways.
Which follows the 3! * 2! = 12

Just rearranging the men we have 6 arrangements...
[w1 m1 w2 m2 w3 m3 w1]
[w1 m2 w2 m1 w3 m3 w1]
[w1 m3 w2 m2 w3 m1 w1]
[w1 m1 w2 m3 w3 m2 w1]
[w1 m2 w2 m3 w3 m1 w1]
[w1 m3 w2 m1 w3 m2 w1]
Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 79 [0], given: 0

### Show Tags

10 Nov 2004, 12:48
IN The SAME way.. ( going by gayathris logic and oxons)... like i said earlier... its 7!*6!... the choices have to be wrong....
_________________

Let's get it right!!!!

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 40

Kudos [?]: 429 [0], given: 0

### Show Tags

10 Nov 2004, 16:09
I just want to say that I totally agree with Venksune's approach and I also came up with 7!*6!
_________________

Best Regards,

Paul

Intern
Joined: 08 Nov 2004
Posts: 46
Location: Montreal
Followers: 0

Kudos [?]: 23 [0], given: 0

### Show Tags

10 Nov 2004, 21:28
You're right guys... I got this from the following link, and it seems the answer choices are wrong

Senior Manager
Joined: 25 Nov 2011
Posts: 261
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 4

Kudos [?]: 167 [0], given: 20

### Show Tags

19 Feb 2012, 04:34
ComplexVision wrote:
You're right guys... I got this from the following link, and it seems the answer choices are wrong

Yes, answer choices should be definitely wrong. I also agree with 6!7!
_________________

-------------------------
-Aravind Chembeti

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13547
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

20 Sep 2013, 00:07
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 Club Legend
Joined: 09 Sep 2013
Posts: 13547
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

13 Feb 2015, 05:59
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.
_________________
Intern
Joined: 31 Oct 2015
Posts: 37
Followers: 0

Kudos [?]: 2 [0], given: 53

Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

31 Dec 2015, 17:47
This is a very nice question.

If the men and women were seated in a line not a circle, the answer would be 7!*7!.

However, because they are seated in a circle, where the is no beginning or end, we have to account for repetitions of a pattern that occur when the head to tail arrangement duplicates another pattern but at a different starting point on the circle. For any one pattern, composed of 7 entities, one would expect seven duplicates with different starting points on the circle. Therefore to reach the correct answer one should account for this repetition by dividing 7!*7! by 7 = 7!*6!.

For further clarification consider this:

In a linear arrangement. 1234567 and 2345671 are different patterns. However, in a circular arrangement they are one and the same. So are 3456712, 4567123, 5671234, 6712345 and 7123456.

Because the men and women alternate positionso on the table the pattern advances by two at a time, so even though there are 14 people on the table, there are only 7 positions in each pattern.
Manager
Joined: 22 Jan 2014
Posts: 138
WE: Project Management (Computer Hardware)
Followers: 0

Kudos [?]: 54 [0], given: 135

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

03 Jan 2016, 02:26
ComplexVision wrote:
Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many ways can that be done?

A. 5!*6!
B. 6!*6!
C. 5!*7!
D. 6!*7!
E. 7!*7!

there are 14 seats on the circular table. for no woman to sit together, the arrangement would be -> MWMWMWM....

so, seat the 7 men at the circular table first in (7-1)! = 6!
now the seats for each woman would be between 2 men, hence making it a linear arrangement which can be done in 7!

answer = 6!*7! = option d.
_________________

Illegitimi non carborundum.

Intern
Joined: 31 Oct 2015
Posts: 37
Followers: 0

Kudos [?]: 2 [0], given: 53

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

03 Jan 2016, 05:28
It doesn't matter if you're starting with a man or a woman. If they are alternating in a circular arrangement, two women will never be seated next to each other. My answer was just an explanation of why permutations in a circular arrangement = (n-1)! instead of n!. The group of 14 doesn't literally shift position down the table, but an arbitrary starting point of every set of seven permutations of a sequence, such as the set shown in my answer above, seems to move up and down the circle as you shift between these permutations but the sequence itself remains unchanged.
Intern
Status: Vice President
Joined: 16 May 2016
Posts: 14
Location: India
Concentration: Finance, Strategy
GMAT 1: 700 Q48 V38
GPA: 2
WE: Operations (Other)
Followers: 0

Kudos [?]: 0 [0], given: 115

Re: Seven men and seven women have to sit around a circular [#permalink]

### Show Tags

13 Aug 2016, 09:53
37 seconds. Just make any one group sit in a circle first. Making people sit in circle has (n-1)! arrangements.
Now when people sit in circle, the gaps between them are equal to the number of people. So make the other gender group sit in those gaps so that no same gender sits adjacent.
n!*(n-1)!

D
_________________

Route to 700+

Re: Seven men and seven women have to sit around a circular   [#permalink] 13 Aug 2016, 09:53
Similar topics Replies Last post
Similar
Topics:
3 A division of a company consists of seven men and five women 6 08 Jan 2013, 09:42
12 Seven family members are seated around their circular dinner 7 04 Oct 2010, 05:30
27 Seven men and five women have to sit around a circular table 30 28 Jul 2010, 18:53
24 Seven men and seven women have to sit around a circular 12 07 Apr 2010, 15:10
31 A group of four women and three men have tickets for seven a 12 30 Dec 2009, 20:43
Display posts from previous: Sort by