Find all School-related info fast with the new School-Specific MBA Forum

It is currently 03 May 2016, 15:27
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How about some round table party? In how many ways can 4 men

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
avatar
Joined: 24 Sep 2005
Posts: 6
Followers: 0

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

How about some round table party? In how many ways can 4 men [#permalink]

Show Tags

New post 24 Sep 2005, 18:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How about some round table party?

In how many ways can 4 men and 4 women sit at a round table so that no two men are adjacent?

Please, explain you answer.
Director
Director
avatar
Joined: 23 Jun 2005
Posts: 847
GMAT 1: 740 Q48 V42
Followers: 5

Kudos [?]: 48 [0], given: 1

 [#permalink]

Show Tags

New post 25 Sep 2005, 10:31
Is it 16? I'll explain if it's right.
VP
VP
avatar
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 72 [0], given: 0

 [#permalink]

Show Tags

New post 25 Sep 2005, 11:48
4 men can sit at 4 seats in:
4P4 = 4! = 24 ways

now consider one position between each two adjecent men. 4 women need to sit in these 4 spots in order for condition to be valid (no two men together). this can, again, be done in:
4P4 = 24 ways.

total arrangments = 24 * 24
Intern
Intern
avatar
Joined: 27 Aug 2005
Posts: 33
Followers: 0

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

 [#permalink]

Show Tags

New post 25 Sep 2005, 12:14
Assuming that the round table has 8 seats:

If 1st seat is taken by a man, then we have 1,3,5,7 seats occupied by men and 2,4,6,8 occupied by women

So for the 4 men, we have 4P4 ways i.e., = 24
and for the 4 women, we have 4P4 ways i.e., 24 ways

so for 4 men and 4 women sitting such that no 2 men are together, we have 24 * 24 ways.

But, the 4 men can also sit at 2,4,6,8 seats and women can sit at 1,3,5,7 seats.

So, The total = 24 * 24 + 24*24 = 2(24*24) ways.
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 36 [0], given: 0

 [#permalink]

Show Tags

New post 25 Sep 2005, 19:02
I also found 2*24*24

in order to have 4 women and 4 men together at a table without adjecent men, there is 1 general layout but in reality 2 configuration : MWMWMWMW or WMWMWMWM

4 positions for men and they are 4 so : 4! ways (=24)
4 positions for women and they are 4 so : 4! ways (=24)

so total 2*24*24 ways
Intern
Intern
avatar
Joined: 24 Sep 2005
Posts: 6
Followers: 0

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

 [#permalink]

Show Tags

New post 25 Sep 2005, 21:10
According to the authors of this case: Number of circular permutations:
4! (4-1)!= 4!*3! =144

Honestly, I spent more than 2 min to figure that out.

Correct me if I am wrong: In case we have 4 women, 4 men and 4 children:

4! * 4! * 3! and so on.
_________________

It isn't the mountains ahead that wear you out, it's the grain of sand in your shoe.

VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 36 [0], given: 0

 [#permalink]

Show Tags

New post 26 Sep 2005, 02:25
redneckboy,

I dont get it clearly...

how can it be this ? :(

why not 4!*4!*2
Senior Manager
Senior Manager
avatar
Joined: 29 Nov 2004
Posts: 484
Location: Chicago
Followers: 1

Kudos [?]: 14 [0], given: 0

 [#permalink]

Show Tags

New post 26 Sep 2005, 06:50
I get 144,

Four ladies with one seat between them and the men can be arranged in the inbetween seats

4! + 4! = 48 ways

the four ladies can shift one seat, two more time, the fourth time they will be back to duplication

so 48+48+48 = 144
_________________

Fear Mediocrity, Respect Ignorance

Intern
Intern
User avatar
Joined: 05 Sep 2005
Posts: 11
Followers: 0

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

 [#permalink]

Show Tags

New post 26 Sep 2005, 07:03
My answer is also 4!*4!/4

Take this problem as an example: Arrange 3 men and 3 women to a round table so that no two men are adjacent.

There are 2*3! way to arrange the men to the table (to 1st, 3rd and 5th AND to 2nd, 4th and 6th) (let's call these 3 men A, B and C)
After arranging the men, there are 3! way to arrange the women to the table (call them 1, 2 and 3)

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!

Similarly, the answer for the posted question is 4!*(4-1)!


Hope this help!

Tibeo - Vietnam
_________________

I can, therefore I am

SVP
SVP
User avatar
Joined: 28 May 2005
Posts: 1723
Location: Dhaka
Followers: 7

Kudos [?]: 164 [0], given: 0

 [#permalink]

Show Tags

New post 26 Sep 2005, 08:04
This a massed up one....
_________________

hey ya......

VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 36 [0], given: 0

 [#permalink]

Show Tags

New post 26 Sep 2005, 18:39
tibeo - vietnam wrote:

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!



Sorry, I still try to get this one :?
We are talking about arragement , right ? It should be permutations no ? I don't understand why you need to divide by 6 ?
For me the total number of possibilities is 2*3!*3!
If someone can explain please :(
Intern
Intern
avatar
Joined: 19 Aug 2005
Posts: 41
Followers: 0

Kudos [?]: 3 [0], given: 0

 [#permalink]

Show Tags

New post 26 Sep 2005, 23:57
yeah... makes no sense at all
Intern
Intern
avatar
Joined: 25 Jun 2005
Posts: 23
Location: Bay Area, CA
Followers: 0

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

Re: circular permutation [#permalink]

Show Tags

New post 02 Oct 2005, 23:11
redneckboy wrote:
How about some round table party?

In how many ways can 4 men and 4 women sit at a round table so that no two men are adjacent?

Please, explain you answer.


first man can sit in 4 ways
first woman can sit in 4 ways

second man - 3 ways
second woman - 3 ways

third man - 2 ways
third woman - 2 ways

4th man - 1 seat left
4th woman - 1 seat left

total ways - 4*4*3*3*2*2*1*1 = 576
_________________

If you can't change the people, change the people.

Manager
Manager
User avatar
Joined: 03 Aug 2005
Posts: 134
Followers: 1

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

 [#permalink]

Show Tags

New post 03 Oct 2005, 01:35
I think there are two possible interpretations for this question.

If you consider that we are being asked only about the order of the people, that is, with a table of four: ABCD = BCDA the answer is 4!*3!

If you consider those cases different, the answer is 4!*4!
Senior Manager
Senior Manager
avatar
Joined: 04 May 2005
Posts: 282
Location: CA, USA
Followers: 1

Kudos [?]: 35 [0], given: 0

 [#permalink]

Show Tags

New post 03 Oct 2005, 17:38
got this circular permutation link:
http://mathworld.wolfram.com/CircularPermutation.html

arrange men first, it is a circular permutation problem, with
3! ways, then arrange women, have 4! ways. Since they are
independent, total is 4! x 3!
Senior Manager
Senior Manager
avatar
Joined: 09 Aug 2005
Posts: 285
Followers: 1

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

 [#permalink]

Show Tags

New post 22 Jan 2006, 22:06
redneckboy wrote:
According to the authors of this case: Number of circular permutations:
4! (4-1)!= 4!*3! =144

Honestly, I spent more than 2 min to figure that out.

Correct me if I am wrong: In case we have 4 women, 4 men and 4 children:

4! * 4! * 3! and so on.


OK guys,

How about this

total ways in which 8 people can sit together = (8-1)!

if two men sit side my side = (7-1)! x2

if three men sit side my side = (6-1) x 3P3

if 4 men sit next to each other = (5-1) x 4P4

therefore our answer is

(8-1)! - ( 6! x 2 ) - (5! x 6) - (4! x 24)

5760 - 1440 - 600 - 576 = 3144

how do you feel about this approach - please help me understand any flawed concept.

Thank you
Director
Director
avatar
Joined: 26 Sep 2005
Posts: 576
Location: Munich,Germany
Followers: 1

Kudos [?]: 15 [0], given: 0

 [#permalink]

Show Tags

New post 25 Jan 2006, 03:07
my vote is for 4!*4!*2
  [#permalink] 25 Jan 2006, 03:07
Display posts from previous: Sort by

How about some round table party? In how many ways can 4 men

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.