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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]

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28 May 2010, 04:04

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

Schools: INSEAD - dinged; IE-admit; ISB - admit; IIMB-admit; SPJain- admit; IIM C- Admit, IIM A - Dinged after interview. Finally joining IIM B

WE 1: 3

WE 2: 2

WE 3: 2

Re: please help with the Seating arrangement problems [#permalink]

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28 May 2010, 07:49

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

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]

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01 Aug 2010, 00:28

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]

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02 Sep 2013, 04:06

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]

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14 Oct 2013, 21:06

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: \(5!6!\).

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]

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09 Jan 2015, 01:48

1

This post was BOOKMARKED

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!?

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]

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03 Jul 2016, 12:20

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