Problem Solving

At some point in circular permutation you start with a man and end ordering with the man. If you order clock-wise you have A B C D E F, but if you order counter-clock wise you have F E D C B A - your beginning for one ordered set becomes an end for the other ordered set made in different direction, depends on how you look at the order - and now you need to discount by one row for the circular (symmetric) nature ordering. You account for this by discounting (6-1)! However, you bring up 6P4 and make repeating orders like the one highlighted in red.

If you had six women you could start with a man and end with a woman. Your ordered set then would be A B C D E F - Some sixth woman (other women are assumed allocated between men, not shown/typed). Your reversed order would be Some sixth woman - F E D C B A. Because you always order the men first, counter-clock wise you would start ordering from man and not from Some sixth woman. This is the case with six men and six women.

Here we have six men and four women - we may start ordering with men and end ordering with men, which will make our clock and counter-clock wise ordering repeating.

One more issue, if you look at the way 5!4!*4P2 you could inquire whether two men (residual men) are allowed to have one woman between them. Yes, they are allowed because the order of men is permuted each time and these two men will be different always. All men will have women following them.

5!*6P4 is correct. Here is another way to do it, where the women are seated first. It's not as good as the solution already presented, but it can be instructive to look at other approaches.

Seat the women around the table first. This can be done in 3! ways. Now, take any four men and place one between each pair of women to satisfy the restriction. Don't worry about the order yet. Now we have 4 women and 4 men alternating around the table, and we have to place the two remaining men somewhere. There are 4 men seated at the table, and because they are arranged in an alternating pattern, each of the new men MUST be placed next to one of the existing 4 men. Now, the two remaining men may both be placed next to the same man, which could be done in 4 ways, or we could choose 2 different men to place the 2 new men next to, which can be done in 4C2 ways. So, there are 4+4C2=10 ways to do this. Now, the men can be rearranged among themselves in 6! ways. So, putting it all together, there are 3!*10*6! ways to arrange the people that satisfy the restriction. It can easily be verified that this is equivalent to 5!*6P4.

@pemdas: how are your two examples that were highlighted in red identical if everyone is not sitting between the same two people in both cases? For example, in the first ordering, Y sits between C and D, and in the second one, Y sits between C and B.

GmatMathPro wrote: @pemdas: how are your two examples that were highlighted in red identical if everyone is not sitting between the same two people in both cases? For example, in the first ordering, Y sits between C and D, and in the second one, Y sits between C and B.

you mean this

the assignment of woman set such as A-V B-X C-Y D-Z E F is the same as F E D-Z C-Y B-X A-V in circular order

if you read from the right to the left, these are the same. I kept C-Y pattern to show similarity, indeed they are A-V B-X C-Y D-Z E F is the same as F E Z-D Y-C X-B V-A. The whole point is that the beginning of one becomes the end of the other in counter-clock order (as usually in circ. perms)

in your solution I didn't get 3!, perhaps you discounted the number of arranged orders (permutations) of women by one raw, but wait you don't need to. The symmetry is not around women here, their number is less anyways, the symmetry will be around men, with more emphasize even after you put additional men on the left and right sides. Why you start with 3! then?

pemdas wrote:the assignment of woman set such as A-V B-X C-Y D-Z E F is the same as F E D-Z C-Y B-X A-V in circular order. if you read from the right to the left, these are the same. I kept C-Y pattern to show similarity, indeed they are A-V B-X C-Y D-Z E F is the same as F E Z-D Y-C X-B V-A. The whole point is that the beginning of one becomes the end of the other in counter-clock order (as usually in circ. perms)

Okay, but these would still not be identical because even though everyone is sitting between the same two people, they don't have the same people on their right and left. If I understand your notation correctly (and please correct me if I don't), the 2 orders you specify would look like this if they were placed around a circular table:

F E Z-D Y-C X-B V-A

A-V B-X C-Y D-Z E F


So, for example, in the first one, A has V on his right and F on his left, but in the second one, A has V on his left and F on his right. These arrangements are not considered equivalent in this context. If they were, there would only be one way to arrange three people around a circular table, but according to the formula (n-1)!, there are 2.

in your solution I didn't get 3!, perhaps you discounted the number of arranged orders (permutations) of women by one raw, but wait you don't need to. The symmetry is not around women here, their number is less anyways, the symmetry will be around men, with more emphasize even after you put additional men on the left and right sides. Why you start with 3! then?

I used 3! because I was seating 4 women around around a circular table, which can be done in (4-1)! ways. I'm not sure what you mean by the symmetry isn't around women. It doesn't matter whether we place the women or the men first, even though placing the women first makes it a bit more complicated. Once we fix the relative positions of the women (or men) around the table, we no longer have to take the circular symmetries into account because we have some points of reference.

@Pete, I am sure if we list the sets we will end up in repeating series within 5!*6P4

as per the identical sets - again, I mean that one set A->F reversed F<-A is the the same as F<-A and A->F. One's beginning clock-wise is the same as one's ending counter-clock wise, the concept of circ. permutations

And in your case you arrange the women first 3! Then you are taking each ordered set out of the discounted by one raw (4-1)! sets of women (circ. perms are adjusted) and for each set of women you have 4! men and 10 ... why you start with women? The concept of circ. perms insist on discounting by one raw only if you start and end with the same element in two ways - clock wise and counter-clock wise

I can't get 3! start up

pemdas wrote:@Pete, I am sure if we list the sets we will end up in repeating series within 5!*6P4

This gives me an idea....

Okay, so we at least both agree that the number of arrangements is in the tens of thousands, so clearly listing them is unfeasible. However, what if we look at a simpler case: 4 men and 2 women around a circular table, and the women can't sit together. Using the method originally suggested by user123321, it would be 3!*4P2=72. I went ahead and made a list of the 72 possible arrangements in the following chart. Picture the first person at the head of the table (let's say due north), and the other people arranged clockwise around the table in sequence. The men are A,B,C, and D. The women are Y and Z. To deal with the circular repeats, I fixed 'A' so that he is always at the head of the table.



You seem to disagree with this method, so could you point out which of these sequences are either invalid or repeats. Or, if you agree that the answer is 72, can you explain why you think the method works for 4 men and 2 women, but not 6 men and 4 women?

Nice!

GmatMathPro wrote:

pemdas wrote:@Pete, I am sure if we list the sets we will end up in repeating series within 5!*6P4

This gives me an idea....

Okay, so we at least both agree that the number of arrangements is in the tens of thousands, so clearly listing them is unfeasible. However, what if we look at a simpler case: 4 men and 2 women around a circular table, and the women can't sit together. Using the method originally suggested by user123321, it would be 3!*4P2=72. I went ahead and made a list of the 72 possible arrangements in the following chart. Picture the first person at the head of the table (let's say due north), and the other people arranged clockwise around the table in sequence. The men are A,B,C, and D. The women are Y and Z. To deal with the circular repeats, I fixed 'A' so that he is always at the head of the table.



You seem to disagree with this method, so could you point out which of these sequences are either invalid or repeats. Or, if you agree that the answer is 72, can you explain why you think the method works for 4 men and 2 women, but not 6 men and 4 women?

pemdas

Just try the same question with linear arrangements, instead of circular arrangements. Then apply the same concept to circular arrangements. Then it will be easy to understand.

user123321

Wow! Clash of Titans

So, let me conclude the post by summarizing:

Circular Arrangement: Therefore 6 men can sit: 6-1-5! way

Then 4 women can sit anywhere between then because they don't want to sit together...Cool...
Since order is required, women can sit 6P4 ways.

Ans = 5!*6P4
Please confirm if my take way is right?

my takeaway:
For alternating circular permutations with two sets of elements N and n where N>=n, we need to fix N around the circle such as (N-1)!, then combine it with the arranged order of n elements with N options -> NPn

No repeats must be sought as long as the circular permutation order has been fixed with the first set.

question:
How can six men and four women sit at the table if only two women should sit next to each other? 5!*6C4*4P2

first, fix the circular permutation of men, 5!
make the ordered arrangement for two women placed with six available options, 6C4 (leaving out two options for other arrangements)*4P2 (combining two ordered arrangements within four options)

>> stick two women for placing them next to each other

@Pete, what's your opinion?

pemdas wrote:question:
How can six men and four women sit at the table if only two women should sit next to each other? 5!*6C4*4P2

Permutation for 6 men => 5!

only two women should sit next to each other..
so first we select the two women who sit together == 4P2 ways
we take the two women as a single entity
so now we have 6 places for 3 ==> 6P3

So the no of arrangements should be 5!*6P3*4P2

It should be noted that the chances of seeing this problem on the GMAT are pretty much nil. Questions about circular permutations are rare. I've never seen or heard of an actual GMAT question that asked about a circular permutation WITH RESTRICTIONS.

LalaB wrote:A group of ten people (six men and four women) wants to sit in a circular table, and no 2 women want to seat together. in how many different ways can they seat down?

Dear Experts/ fellow members

Below is the method i approached, but the answer is not the same. I am sure i am counting some extra cases here but couldnt figure out which one. So here goes my attempt at it...

We have M M M M M M W W W W

Total number of ways of arranging 10 people on circular table --- 10-1= 9! ways

Now if we consider the case where all women will sit together and minus it from the total number of ways ,, that should be equal to saying that no 2 women sit together.

So for this we consider W W W W as one person and now we have M M M M M M WWWW.
These 7 can be arranged in circle in 6!*4!ways, 4!here signifies that all women can interchange their position.

So as per the logic total number of ways should be 9! - 6!*4! which is not equal to the correct answer above decided in majority.

Please correct me where i m going wrong. Does my way of doing it includes the cases where 2/3 women can sit together but not all women can sit together.????

thanks for contributing, I guess the q. stem is ambiguous. I meant only women sit by two; you inferred two women sit together and the rest can be permuted. I tried to follow your explanations, you have seated two women next to each other and the other two women are ordered as such, in total three elements two women joint, one women and one women, 6P3 correct?

The only thing, I was concerned with why you have permuted here 6P3, don't you need the number of options for three seats? You are permuting among 4 (4P2) anyways, why you permute 6P3? Doesn't this add to the additional sets with other 6 men permuted already within the circular permutation (5!)? Would you reconsider, 6C3 instead?

rijul007 wrote:

pemdas wrote:question:
How can six men and four women sit at the table if only two women should sit next to each other? 5!*6C4*4P2

Permutation for 6 men => 5!

only two women should sit next to each other..
so first we select the two women who sit together == 4P2 ways
we take the two women as a single entity
so now we have 6 places for 3 ==> 6P3

So the no of arrangements should be 5!*6P3*4P2

pemdas wrote:thanks for contributing, I guess the q. stem is ambiguous. I meant only women sit by two; you inferred two women sit together and the rest can be permuted. I tried to follow your explanations, you have seated two women next to each other and the other two women are ordered as such, in total three elements two women joint, one women and one women, 6P3 correct?

Before I jump in here, let me make sure I understand. In the new question you're posing, you want the four women to be seated in pairs so that each woman is seated next to exactly one woman? Is that right?