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

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

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27 May 2010, 21:30
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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.
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ASHISH DONGRE
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28 May 2010, 03:02
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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.
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28 May 2010, 00:29
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...
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28 May 2010, 01:18
1
I am providing my thinking below.
Seating arrangement in a row = 6! * 5!
Seating arrangement in a table = 5! * 5!
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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
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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.

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28 May 2010, 08:46
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).
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31 Jul 2010, 03:56
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?
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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: $$5!6!$$.

Hope it helps.

It's really clear. Thanks for new knowledge!
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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.
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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: $$5!6!$$.

Hope it helps.

Hi Banuel! Why the solution is not 5!*6*5!?
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15 Oct 2013, 09:52
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?
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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
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!?
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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

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23 Dec 2017, 03:33
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
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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13 Jun 2018, 23:36
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

In the second case mentioned by you, as we have now 6 men and 6 women, why aren't we multiplying by 2!?

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: $$5!6!$$.

Thanks
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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13 Jun 2018, 23:53
@s 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

In the second case mentioned by you, as we have now 6 men and 6 women, why aren't we multiplying by 2!?

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: $$5!6!$$.

Thanks

Why should we multiply by 2!?
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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14 Jun 2018, 05:29
Hi Bunuel

In the second case mentioned by you, as we have now 6 men and 6 women, why aren't we multiplying by 2!?

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: $$5!6!$$.

Thanks[/quote]

Why should we multiply by 2!?[/quote]

because men and women can interchange their place.
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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14 Jun 2018, 13:37
To count circular arrangements without applying a special formula:
1. Place someone in the circle.
2. Count the number of ways to arrange the REMAINING people.

Quote:
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?

Here, men and women must ALTERNATE.
After one of the 6 men has been placed at the table, count the number of options for each empty seat, moving clockwise around the table:
Number of options for the first empty seat = 6. (Any of the 6 women.) )
Number of options for the next empty seat = 5. (Any of the 5 remaining men.)
Number of options for the next empty seat = 5. (Any of the 5 remaining women.)
Number of options for the next empty seat = 4. (Any of the 4 remaining men.)
Number of options for the next empty seat = 4. (Any of the 4 remaining women.)
Number of options for the next empty seat = 3. (Any of the 3 remaining men.)
Number of options for the next empty seat = 3. (Any of the 3 remaining women.)
Number of options for the next empty seat = 2. (Either of the 2 remaining men.)
Number of options for the next empty seat = 2. (Either of the 2 remaining women.)
Number of options for the next empty seat = 1. (Only 1 man left.)
Number of options for the last empty seat = 1. (Only 1 woman left.)
To combine these options, we multiply:
6*5*5*4*4*3*3*2*2*1*1 = 6!5!.
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Re: The number of ways in which 5 men and 6 women can be seated  [#permalink]

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Re: The number of ways in which 5 men and 6 women can be seated   [#permalink] 03 Jul 2019, 23:09
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