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

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Intern
Joined: 24 Sep 2005
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How about some round table party? In how many ways can 4 men  [#permalink]

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24 Sep 2005, 17:45
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?

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Director
Joined: 23 Jun 2005
Posts: 783
GMAT 1: 740 Q48 V42

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25 Sep 2005, 09:31
Is it 16? I'll explain if it's right.
VP
Joined: 22 Aug 2005
Posts: 1073
Location: CA

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25 Sep 2005, 10:48
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
Joined: 27 Aug 2005
Posts: 33

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25 Sep 2005, 11:14
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
Joined: 13 Jun 2004
Posts: 1074
Location: London, UK
Schools: Tuck'08

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25 Sep 2005, 18:02
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
Joined: 24 Sep 2005
Posts: 6

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25 Sep 2005, 20:10
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.
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VP
Joined: 13 Jun 2004
Posts: 1074
Location: London, UK
Schools: Tuck'08

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26 Sep 2005, 01:25
redneckboy,

I dont get it clearly...

how can it be this ?

why not 4!*4!*2
Senior Manager
Joined: 29 Nov 2004
Posts: 443
Location: Chicago

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26 Sep 2005, 05:50
1
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
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Intern
Joined: 05 Sep 2005
Posts: 9

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26 Sep 2005, 06:03
2

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
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SVP
Joined: 28 May 2005
Posts: 1584
Location: Dhaka

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26 Sep 2005, 07:04
This a massed up one....
VP
Joined: 13 Jun 2004
Posts: 1074
Location: London, UK
Schools: Tuck'08

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26 Sep 2005, 17:39
1
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!
Intern
Joined: 19 Aug 2005
Posts: 41

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26 Sep 2005, 22:57
1
yeah... makes no sense at all
Intern
Joined: 25 Jun 2005
Posts: 23
Location: Bay Area, CA

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02 Oct 2005, 22:11
1
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?

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
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Manager
Joined: 03 Aug 2005
Posts: 128

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03 Oct 2005, 00:35
1
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
Joined: 04 May 2005
Posts: 265
Location: CA, USA

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03 Oct 2005, 16:38
2
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
Joined: 09 Aug 2005
Posts: 269

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22 Jan 2006, 21:06
1
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,

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

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

5760 - 1440 - 600 - 576 = 3144

Thank you
Director
Joined: 26 Sep 2005
Posts: 534
Location: Munich,Germany

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25 Jan 2006, 02:07
1
my vote is for 4!*4!*2
Non-Human User
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Re: How about some round table party? In how many ways can 4 men  [#permalink]

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06 Jul 2018, 07:26
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Re: How about some round table party? In how many ways can 4 men &nbs [#permalink] 06 Jul 2018, 07:26
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