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 350,000 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:

Re: please help with the Seating arrangement problems [#permalink]
28 May 2010, 02:02

1

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Re: please help with the Seating arrangement problems [#permalink]
28 May 2010, 03:04

Row = 5! * 6! Circle = 0 (men can be separated from each other with a woman in between but the 6th woman invariably sits next to the 1st woman) Pls correct my understanding if its wrong

Re: please help with the Seating arrangement problems [#permalink]
28 May 2010, 06:49

Thanks all and Bunuel. That is the OA.

I have a doubt still.. Wont we multiply 6!5! by 2!, as there are two possibilities, one we fix women in first position and other that we fix a man in hte first position?

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Hope it helps.

_________________

ASHISH DONGRE BE KIND & GENEROUS TO SHARE THE KUDOS...THE MORE YOUR GIVE THE MORE YOU GET

Re: please help with the Seating arrangement problems [#permalink]
28 May 2010, 07:46

Expert's post

meghash3 wrote:

Thanks all and Bunuel. That is the OA.

I have a doubt still.. Wont we multiply 6!5! by 2!, as there are two possibilities, one we fix women in first position and other that we fix a man in hte first position?

It's only possible woman to be first and than man: W-M-W-M-W-M-W-M-W-M-W (remember there are 6 women and 5 men). _________________

Re: please help with the Seating arrangement problems [#permalink]
31 Jul 2010, 23:28

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Re: The number of ways in which 5 men and 6 women can be seated [#permalink]
02 Sep 2013, 03:06

Bunuel, is my thinking correct for No2?

total: Number of possible sitting arrangements in a circular table: 10! (11-1)!

restriction: Taking 2 men as a set who are not allowed to sit next to each other: 2! Taking 2 women as a set who are not allowed to sit next to each other: 2! The rest of the sitting arrangements is 6! (7-1)!

So, total - restriction = 10! - (2! + 2! + 6!) = 0.

Re: please help with the Seating arrangement problems [#permalink]
14 Oct 2013, 20:06

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Re: please help with the Seating arrangement problems [#permalink]
15 Oct 2013, 08:52

Expert's post

Magdak wrote:

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Hope it helps.

Hi Banuel! Why the solution is not 5!*6*5!?

Can you please elaborate what you mean? _________________

Re: The number of ways in which 5 men and 6 women can be seated [#permalink]
09 Jan 2015, 00:48

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Hope it helps.

Hi Bunuel,

If the questions 1 says in how many different ways 6 men and 6 women can be seated in a row, so that no two women or two men sit together, is the answer: 2x 6! x 6!?

Re: The number of ways in which 5 men and 6 women can be seated [#permalink]
09 Jan 2015, 02:03

Expert's post

vietnammba wrote:

Bunuel wrote:

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is 6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is 5!. So total 6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in (6-1)!=5! # of ways (# of circular permutations of n different objects is (n-1)!). 6 women between them can be seated in 6! # of ways. Total: 5!6!.

Hope it helps.

Hi Bunuel,

If the questions 1 says in how many different ways 6 men and 6 women can be seated in a row, so that no two women or two men sit together, is the answer: 2x 6! x 6!?

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

Almost half of MBA is finally coming to an end. I still have the intensive Capstone remaining which started this week, but things have been ok so far...