Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

# of arrangements of 7 men around a table is \((7-1)!=6!\); There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is \(C^5_7=21\); # of arrangements of 5 women in these slots is \(5!\);

So total: \(6!*21*5!=1,814,400\).

Answer: 1,814,400.

jakolik please post PS questions in PS forum and also try to provide answer choices.
_________________

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

Dear Bunnel

the qs only says that the Women should not sit next 2 each other... but men can right? so y are we assuming

M_M_M_M_M_M_M_ : 14 Places

we can also have:

M_M_M_M_M_MMM: 12 places MMM_M_M_M_M_M: 12 places

in this case the answer will be:

6!*5! right??

Not sure I can follow you...

Those 12 people are around a circular table. So, when no 2 women are together, at least 2 men must be together.

yes, so in this case the solution will be 6! * 5! right?? how is this different from your solution.... our purpose of 2 women not being together can be solved like this also?? basically where did you get 21 from in 6! * 5! * 21 solution provided by you for the above problem

The question clearly says that we have to arrange people on a round table such that no two women are together. We approach this question like this:

We first arrange 7 men on the table which can be done is (n-1)! ways = 6! ways.

Now between these 7 men there are 7 places where women can be placed, we select 5 out of these 7 please which can be done in 7C5 = 7!/5!2! = 21 ways

These 5 women can be arranged in 5! ways.

Thus, we get the solution as 6!*21*5!

I understood this. But what i am trying to ask is that the qs is not saying that 2 men cannot be together. can we have an arrangement like below?[/quote]

Yes, we can definitely arrange it in the way you presented. The question never put a restriction on how men should be arranged. In fact if you see your diagram, you are choosing any 5 of 7 available slots for women. The 5 slots you choose leaves 3 men together.. Hope it makes sense..

Re: Seven men and five women have to sit around a circular table [#permalink]

Show Tags

18 May 2014, 08:03

2

This post received KUDOS

JusTLucK04 wrote:

Awesome approach Bunuel... Why is this wrong neways: 12 Guys..No of ways 11! Take 2 Women and bind them together...Now 11 guys and no of arrangements is 10!*2! This covers all arrangements with at least 2 women together....

No of arrangements for no women together is: 11!- 10!*2! ---> wrong answer Why?

The error is in 10!*2!. This is not taking into account which 2 women are together. This is always done when you know that which two persons are together. For example, if a question says that A & B should always be together, then only you can do it. Now they are not specifying any particular group of women. Hence, this approach will lead you to a wrong answer. Had there been only 2 women in the above question, then your approach would have worked well. If we had 7 men and two women and the rest of the question is same, we would get an answer as:

Approach 1: 6! * 7C2 * 2! = 7! *6

Approach 2: 8! - 7!2! = 6*7!

I hope you get my point.

If you try to apply this approach, you have to first fix those two women who are together, which can be done in 5C2 ways. But when you are arranging them, there would be a case when three women W1W2W3 are together which is included multiple number of times. You have to subtract these cases, which is a very tedious process and hence unproductive.

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

Dear Bunnel

the qs only says that the Women should not sit next 2 each other... but men can right? so y are we assuming

M_M_M_M_M_M_M_ : 14 Places

we can also have:

M_M_M_M_M_MMM: 12 places MMM_M_M_M_M_M: 12 places

in this case the answer will be:

6!*5! right??

Not sure I can follow you...

Those 12 people are around a circular table. So, when no 2 women are together, at least 2 men must be together.

yes, so in this case the solution will be 6! * 5! right?? how is this different from your solution.... our purpose of 2 women not being together can be solved like this also?? basically where did you get 21 from in 6! * 5! * 21 solution provided by you for the above problem

No.

Imagine 7 men around a table so that there is an empty slot between any two:

Attachment:

Table.png [ 4.54 KiB | Viewed 4179 times ]

We'll have 7 slots (blue dots) but we have only 5 women. Now, those 5 can take any of the 7 available slots (and in this case no two women will be together). The number of way to choose 5 out of 7 is \(C^5_7=21\).
_________________

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

Dear Bunnel

the qs only says that the Women should not sit next 2 each other... but men can right? so y are we assuming

M_M_M_M_M_M_M_ : 14 Places

we can also have:

M_M_M_M_M_MMM: 12 places MMM_M_M_M_M_M: 12 places

in this case the answer will be:

6!*5! right??

Not sure I can follow you...

Those 12 people are around a circular table. So, when no 2 women are together, at least 2 men must be together.

yes, so in this case the solution will be 6! * 5! right?? how is this different from your solution.... our purpose of 2 women not being together can be solved like this also?? basically where did you get 21 from in 6! * 5! * 21 solution provided by you for the above problem

The question clearly says that we have to arrange people on a round table such that no two women are together. We approach this question like this:

We first arrange 7 men on the table which can be done is (n-1)! ways = 6! ways.

Now between these 7 men there are 7 places where women can be placed, we select 5 out of these 7 please which can be done in 7C5 = 7!/5!2! = 21 ways

Re: Seven men and five women have to sit around a circular table [#permalink]

Show Tags

18 May 2014, 10:38

1

This post received KUDOS

I visualize these arrangements as following:

Attachment:

Roundarrangement.jpg [ 110.03 KiB | Viewed 1429 times ]

_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

It is a tough question, at least for me So let me try, sure that the following is not the most simple way. I first arrange 5 women around the table, there are 5!/5=4! ways. Then I select randomly 5 men and insert them, again randomly, into places between women, there are C_5_7*5! ways. OK, now is the last step, it remains 2 men, and I can insert them to any places as I like. I distinguish 2 cases: - Case 1: 1 set of 3 men together, then I have to insert the 2 remaining men into the same segment limited by 2 consecutive women. There are 10 places for the first man then 3 for the second, meaning a total of 10*3 ways. Nevertheless, it is remarked that considering the equivalent roles of these three men, I will divide the result of this case by 3. - Case 2: 2 sets of 2 men together, then I insert the 2 remaining men into 2 different segments limited by 2 consecutive women. There are 10 places for the first man and then 8 for the second, meaning a total of 10*8 ways. Again, it is remarked that the result of this case should be divided by 4. In conclusion, the total number of ways is: 4!*C_5_7*5!*[10*3/3+10*8/4]=1814400 ways.
_________________

Dear Bunuel, Can you explain why in a circle it is (n-1)!

From Gmat Club Math Book (combinatorics chapter):

"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:

\(R = \frac{n!}{n} = (n-1)!\)"

Check Combinatorics chapter of Math Book (link in my signature).

Think the mathematic working is reasonable. But how do we rationalize it from normal logic? My thinking was:- if no women(W) are to be together, they will have to be partitioned by men(M). So I paired them up in this manner:-

MW MW MW MW MW MM

So the number of ways to arrange the pairs is 6 !

Next to address the pairing:-

In first pair : 7C1 x 5C1 = 7 x 5 = 35 2nd pair : 6C1 x 4C1 = 6 x 4 = 24 3rd : 5 x 3 = 15 4th : 4 x 2 = 8 5th : 3 x 1 = 3 6th :2C1 x 2C1 = 4

Hence total is 6! x (35 + 24 + 15 + 8 +3 + 4) = 64, 080.

Will anyone be able to point out to me where I've gone wrong?

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

# of arrangements of 7 men around a table is \((7-1)!=6!\); There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is \(C^5_7=21\); # of arrangements of 5 women in these slots is \(5!\);

So total: \(6!*21*5!=1,814,400\).

Answer: 1,814,400.

jakolik please post PS questions in PS forum and also try to provide answer choices.

Bunuel

I understand the 6! part for the #ofmen arrangements. However for the #of women arrangements you use 7C5. I see there are 7 slots available and 5 women. - For the women we are essentially using nPr = nCr r! arrangements. - Let me extend the argument - if there were 7 women, then the #of arrangements for women by the nCr r! logic would be 7C7 7!=7!, however it would seem that we should be using the (r-1)!= 6! (similar to men) for the women too. What am I missing. Thanks
_________________

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

# of arrangements of 7 men around a table is \((7-1)!=6!\); There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is \(C^5_7=21\); # of arrangements of 5 women in these slots is \(5!\);

So total: \(6!*21*5!=1,814,400\).

Answer: 1,814,400.

jakolik please post PS questions in PS forum and also try to provide answer choices.

Bunuel

I understand the 6! part for the #ofmen arrangements. However for the #of women arrangements you use 7C5. I see there are 7 slots available and 5 women. - For the women we are essentially using nPr = nCr r! arrangements. - Let me extend the argument - if there were 7 women, then the #of arrangements for women by the nCr r! logic would be 7C7 7!=7!, however it would seem that we should be using the (r-1)!= 6! (similar to men) for the women too. What am I missing. Thanks

So basically you are saying that women are also placed in a circle and we should use \((n-1)!\) formula. The reason we use \((n-1)!\) for circular arrangements is that when we shift all \(n\) objects by one position, we still would have the same arrangement of all objects relative to each other.

But consider different situation: a table with 6 chairs where every second chair is already taken by men. Now, if we place women in empty places we would have one arrangement BUT if we shift these women by one position we would have different arrangement as relative position of 6 changed. So # of different arrangements in this case would be 3! not 2!.

I understand why we use the (n-1)! for the circular case. What I was confused on is this: I guess then that the men get to use the (n-1)! = nPn/n because they go in FIRST. In the case of the women we use the nPr = nCr r! because they go in NEXT, so in this case the order in which group goes in matters. In the original question with 7Men and 5Women, the answer is 6! 7C5 5!, with 7Men and 7Women, the answer would be 6! 7C7 7! <- because the men that are already seated now influence the arrangement of the women (can't use 6! for women). And that would be the same answer whatever order they go in.

- Now back to the case of 7M and 5W, we solved it first seating the Men to get the 6!. Can we solve it FIRST seating the women and then men, it should be the same answer? - What would be the general answer in this case: C chairs M men and W women (where C >= M+W) and no 2 men or women sit next to each other?
_________________

Re: Seven men and five women have to sit around a circular table [#permalink]

Show Tags

05 Jan 2014, 15:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Hi, in order to creat 5 places for 5 women (where two women will not sit togeather) I believe there are three cases , case one :- already explained by Bunuel case two :- when 2 men sit togeather we could choose 2 men out of 7 by 7c2 ways, now consider those 2 men as single entity , 6 men can sit in circular table in 5! ways, no of places for 5 women are 6 so 6c5, and 5! (no of ways by which women can arrange them self) 7c2 * 5! *6c5 * 5! = 21 * 5! * 6! case three :- when three men sit together, still we can have 5 place for 5 women where they will not sit togeather.

my query is Shouldnt we add case 2 and case 3 Bunnel solution

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

# of arrangements of 7 men around a table is \((7-1)!=6!\); There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is \(C^5_7=21\); # of arrangements of 5 women in these slots is \(5!\);

So total: \(6!*21*5!=1,814,400\).

Answer: 1,814,400.

Dear Bunnel

the qs only says that the Women should not sit next 2 each other... but men can right? so y are we assuming

M_M_M_M_M_M_M_ : 14 Places

we can also have:

M_M_M_M_M_MMM: 12 places MMM_M_M_M_M_M: 12 places

in this case the answer will be:

6!*5! right??
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

gmatclubot

Re: Arrangement In a Circle
[#permalink]
17 May 2014, 05:49

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...