Seven men and five women have to sit around a circular table : GMAT Problem Solving (PS)
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# Seven men and five women have to sit around a circular table

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28 Jul 2010, 18:53
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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?
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29 Jul 2010, 01:41
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jakolik wrote:
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: $$6!*21*5!=1,814,400$$.

jakolik please post PS questions in PS forum and also try to provide answer choices.
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17 May 2014, 09:32
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mittalg wrote:
NGGMAT wrote:
jakolik wrote:
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..
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Re: Seven men and five women have to sit around a circular table [#permalink]

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18 May 2014, 08:03
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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.
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Re: Arrangement In a Circle [#permalink]

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17 May 2014, 07:29
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NGGMAT wrote:
Bunuel wrote:
NGGMAT wrote:
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 $$C^5_7=21$$.
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Re: Arrangement In a Circle [#permalink]

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17 May 2014, 07:41
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NGGMAT wrote:
jakolik wrote:
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!
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Re: Seven men and five women have to sit around a circular table [#permalink]

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18 May 2014, 10:38
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I visualize these arrangements as following:
Attachment:

Roundarrangement.jpg [ 110.03 KiB | Viewed 1429 times ]

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Re: Arrangement In a Circle [#permalink]

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29 Jul 2010, 00:57
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.
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Re: Arrangement In a Circle [#permalink]

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29 Jul 2010, 01:42
Thanks Bunuel and will definitely do in future posts
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Re: Arrangement In a Circle [#permalink]

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31 Jul 2010, 07:11
Dear Bunuel,
Can you explain why in a circle it is (n-1)!
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31 Jul 2010, 18:40
abhijeetkarkare wrote:
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).

Hope it's clear.
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Re: Arrangement In a Circle [#permalink]

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01 Aug 2010, 08:22
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?

Thks.
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Re: Arrangement In a Circle [#permalink]

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07 Aug 2010, 14:23
Bunuel , Can you please explain this part

There will be 7 possible places for women between them

M1_ M2_M3_M4_M5_M6_M7 - i counted 6 possible places , where did i miss ?
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07 Aug 2010, 14:26
ichha148 wrote:
Bunuel , Can you please explain this part

There will be 7 possible places for women between them

M1_ M2_M3_M4_M5_M6_M7 - i counted 6 possible places , where did i miss ?

Men are placed around a circular table so there will be one more place between M7 and M1.
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Re: Arrangement In a Circle [#permalink]

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07 Aug 2010, 15:56
Bunuel wrote:
jakolik wrote:
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: $$6!*21*5!=1,814,400$$.

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
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Re: Arrangement In a Circle [#permalink]

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08 Aug 2010, 00:55
mainhoon wrote:
Bunuel wrote:
jakolik wrote:
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: $$6!*21*5!=1,814,400$$.

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!.

Hope it's clear.
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Re: Arrangement In a Circle [#permalink]

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08 Aug 2010, 07:07
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?
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Re: Arrangement In a Circle [#permalink]

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04 Mar 2014, 22:06
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
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Re: Arrangement In a Circle [#permalink]

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17 May 2014, 05:49
Bunuel wrote:
jakolik wrote:
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: $$6!*21*5!=1,814,400$$.

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

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