It is currently 22 Jan 2018, 06:36

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The number of ways in which 5 men and 6 women can be seated

Author Message
TAGS:

### Hide Tags

Manager
Joined: 24 Mar 2010
Posts: 77
Location: Mumbai, India
WE 1: 3
WE 2: 2
WE 3: 2
The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

27 May 2010, 20:30
1
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

50% (01:00) correct 50% (01:38) wrong based on 34 sessions

### HideShow timer Statistics

The number of ways 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.
_________________

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

Manager
Joined: 20 Apr 2010
Posts: 151
Location: I N D I A

### Show Tags

27 May 2010, 23:29
No of way in which 5 men and 6 women can be seated in a row = 6! * 5! as first has to be a woman then man then woman then man n so on...
Manager
Joined: 16 Mar 2010
Posts: 171

### Show Tags

28 May 2010, 00:18
I am providing my thinking below.
Seating arrangement in a row = 6! * 5!
Seating arrangement in a table = 5! * 5!
Math Expert
Joined: 02 Sep 2009
Posts: 43363

### Show Tags

28 May 2010, 02:02
3
KUDOS
Expert's post
11
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: $$5!6!$$.

Hope it helps.
_________________
Intern
Joined: 01 Apr 2010
Posts: 11

### Show Tags

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
Manager
Joined: 24 Mar 2010
Posts: 77
Location: Mumbai, India
WE 1: 3
WE 2: 2
WE 3: 2

### Show Tags

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

Math Expert
Joined: 02 Sep 2009
Posts: 43363

### Show Tags

28 May 2010, 07:46
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).
_________________
Intern
Joined: 21 Jul 2010
Posts: 4

### Show Tags

31 Jul 2010, 02:56
I know this post is a bit old...but it just caught my eye now

I have a Q, Bunuel

Since its 6 W & 5 M in a circular pattern, there will b 6 slots to choose from for the 5 M

Hence it will b $$6C5 X 5! X 5!$$, won't it?
Intern
Joined: 09 Dec 2008
Posts: 27
Location: Vietnam
Schools: Somewhere

### Show Tags

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: $$5!6!$$.

Hope it helps.

It's really clear. Thanks for new knowledge!
Manager
Joined: 26 Feb 2013
Posts: 170
Re: The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

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.
Intern
Joined: 07 Jan 2013
Posts: 26
Location: Poland
GPA: 3.8

### Show Tags

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: $$5!6!$$.

Hope it helps.

Hi Banuel! Why the solution is not 5!*6*5!?
Math Expert
Joined: 02 Sep 2009
Posts: 43363

### Show Tags

15 Oct 2013, 08:52
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?
_________________
Intern
Joined: 24 Jun 2014
Posts: 44
Re: The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

09 Jan 2015, 00:48
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!?
Math Expert
Joined: 02 Sep 2009
Posts: 43363
Re: The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

09 Jan 2015, 02:03
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!?

Absolutely.

Check other Seating Arrangements in a Row and around a Table questions in our Special Questions Directory.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 14182
Re: The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

27 Sep 2017, 18:07
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.
_________________
Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 233
The number of ways in which 5 men and 6 women can be seated [#permalink]

### Show Tags

23 Dec 2017, 02:33
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 man

I have a question

there are 6 women and 5 men, so
6 women sitting next to each other in a row creates 7 empty spots where men can sit as shown by star symbol as under

* W * W * W * W * W * W *

thus, why cannot be the answer, 6! x 7 x 6 x 5 x 4 x 3

of-course I am questioning about "in a row" case, not about "around a table" one
thanks
The number of ways in which 5 men and 6 women can be seated   [#permalink] 23 Dec 2017, 02:33
Display posts from previous: Sort by