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Seven men and seven women have to sit around a circular

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Seven men and seven women have to sit around a circular [#permalink] New post 07 Apr 2010, 15:10
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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?
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Re: Circular probability question? [#permalink] New post 07 Apr 2010, 20:06
7 men can sit around a circular table in (7-1)! ways = 6!
[Logic: no: of asymmetric circular permutations of n objects is (n-1)!.]

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 :wink:
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Re: Circular probability question? [#permalink] New post 07 Apr 2010, 21:10
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...?
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Re: Circular probability question? [#permalink] New post 08 Apr 2010, 06:27
since we have already arranged the 7 men in a circular fashion, the question, thereafter, ceases to be based on circular permutation.

[Tip :idea: :- when you consider such circular scenarios, imagine a "passing-the-parcel" game in either clockwise or anti-clockwise 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.
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Re: Circular probability question? [#permalink] New post 08 Apr 2010, 11:37
I would have gone with 7!*7!... can someone please explain why its 6! and not 7! for the men?
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Re: Circular probability question? [#permalink] New post 09 Apr 2010, 22:05
idiot is right ..
another discussion here ..
http://www.urch.com/forums/gmat-problem ... ility.html
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Re: Circular probability question? [#permalink] New post 20 Jan 2011, 08:47
idiot wrote:
7 men can sit around a circular table in (7-1)! ways = 6!
[Logic: no: of asymmetric circular permutations of n objects is (n-1)!.]

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 :wink:


Are you sure about that??

Why are women not considerad also circular???
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Re: Circular probability question? [#permalink] New post 20 Jan 2011, 20:31
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craky wrote:
idiot wrote:
7 men can sit around a circular table in (7-1)! ways = 6!
[Logic: no: of asymmetric circular permutations of n objects is (n-1)!.]

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 :wink:


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 non-adjacent 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.
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Re: Circular probability question? [#permalink] New post 21 Jan 2011, 11:58
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craky wrote:
idiot wrote:
7 men can sit around a circular table in (7-1)! ways = 6!
[Logic: no: of asymmetric circular permutations of n objects is (n-1)!.]

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 :wink:


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 (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
please-help-with-the-seating-arrangement-problems-94915.html

Hope it helps.
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Re: Seven men and seven women have to sit around a circular [#permalink] New post 21 Sep 2013, 10:51
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Re: Seven men and seven women have to sit around a circular   [#permalink] 21 Sep 2013, 10:51
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