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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 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.
Edited the answer choices and added the OA
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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

I don't have the OA... Please help :wall


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:
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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]

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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

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

http://www.testmagic.com/forums/showthread.php?t=15259
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New post 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 :hammer

http://www.testmagic.com/forums/showthread.php?t=15259


Yes, answer choices should be definitely wrong. I also agree with 6!7!
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New post 13 Feb 2015, 05:59
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New post 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.
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New post 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.
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New post 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.
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Re: Seven men and seven women have to sit around a circular [#permalink]

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New post 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)!

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Re: Seven men and seven women have to sit around a circular   [#permalink] 13 Aug 2016, 09:53
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