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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, 18:45
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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
Joined: 23 Jun 2005
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Is it 16? I'll explain if it's right.



VP
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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



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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.



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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



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According to the authors of this case: Number of circular permutations:
4! (41)!= 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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redneckboy,
I dont get it clearly...
how can it be this ?
why not 4!*4!*2



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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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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!*(31)!
Similarly, the answer for the posted question is 4!*(41)!
Hope this help!
Tibeo  Vietnam
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This a massed up one....
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hey ya......



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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!*(31)!
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



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yeah... makes no sense at all



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Re: circular permutation [#permalink]
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02 Oct 2005, 23:11
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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
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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!



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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!



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redneckboy wrote: According to the authors of this case: Number of circular permutations: 4! (41)!= 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 = (81)!
if two men sit side my side = (71)! x2
if three men sit side my side = (61) x 3P3
if 4 men sit next to each other = (51) x 4P4
therefore our answer is
(81)!  ( 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



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my vote is for 4!*4!*2



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Re: How about some round table party? In how many ways can 4 men [#permalink]
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Re: How about some round table party? In how many ways can 4 men
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16 Jun 2017, 16:51






