Circular Permutation

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Circular Permutation

by knight247 » Thu Sep 22, 2011 10:03 pm
In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?

Here is my take. Lets begin with any group first eg Men. 10 Men can be seated around the table in 9! ways. 10 vacant seats are now left between the men which can be filled by the 10 women in 10! ways. 20 vacant spots are now left around the table and can be filled by the 10 children in 20P10 ways=20!/10!

So total number of ways=9!10!20!/10!=[spoiler]9!20![/spoiler] Is my approach correct?
Last edited by knight247 on Thu Sep 22, 2011 10:35 pm, edited 1 time in total.

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by cans » Thu Sep 22, 2011 10:22 pm
You arranged men and women, but the way you arranged children, men and women won't be alternate..
Men and women: 9! 10!
Now for children, just take one place and arrange them: 20C1*10!
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by shankar.ashwin » Thu Sep 22, 2011 10:28 pm
I think the question is poorly worded;

Are only men and women position alternate; then all children would sit together? or could children sit inbetween men and women such that no 2 men or women sit together? And together there are only 30 seats in the table?

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by knight247 » Fri Sep 23, 2011 12:48 am
ALL have to alternate. So we need to have MWC MWC etc or MCW MCW etc.

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by shankar.ashwin » Fri Sep 23, 2011 2:20 am
I am not sure of my answer;

My guess;
You got 10 groups of 3 people (M W C) in some order in a circular arrangement.

Consider M sits first; 9! arrangements are possible.

For both W and C; they have 10 position and 10 people; so thats 10! each.

Also MWC can be in any order among themselves in a group; so thats 3! ways among themselves.

In total; 9! 10! 10! 3!

P.S In your solution; after all the women are seated, you have only 10 positions left, where does 20P10 for the children come from when women are already seated? Again I am unsure of my answer though. You got an OA?

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by pemdas » Fri Sep 23, 2011 4:02 am
edited: (10-1)!*P(3,1)=9*9!
knight247 wrote:In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?

Here is my take. Lets begin with any group first eg Men. 10 Men can be seated around the table in 9! ways. 10 vacant seats are now left between the men which can be filled by the 10 women in 10! ways. 20 vacant spots are now left around the table and can be filled by the 10 children in 20P10 ways=20!/10!

So total number of ways=9!10!20!/10!=[spoiler]9!20![/spoiler] Is my approach correct?
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by gmatclubmember » Fri Sep 23, 2011 5:53 am
Men can be arranged in 9! ways. And the remaining Women and children can be arranged in 10! ways each.
Total no. of combo 9!*10!*10!.

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by Brent@GMATPrepNow » Fri Sep 23, 2011 5:55 am
knight247 wrote:In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?

Here is my take. Lets begin with any group first eg Men. 10 Men can be seated around the table in 9! ways. 10 vacant seats are now left between the men which can be filled by the 10 women in 10! ways. 20 vacant spots are now left around the table and can be filled by the 10 children in 20P10 ways=20!/10!

So total number of ways=9!10!20!/10!=[spoiler]9!20![/spoiler] Is my approach correct?
The question seems pretty ambiguous to me. It's hard to tell whether the physical location (vs the relative location) of each seated person creates a new arrangement.
For example, if we have one arrangement, and then everyone shifts 1 seat to the left, does this create a new arrangement, even though the relative location of each person is the same?

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by gmatclubmember » Fri Sep 23, 2011 5:55 am
shankar.ashwin wrote:I am not sure of my answer;

My guess;
You got 10 groups of 3 people (M W C) in some order in a circular arrangement.

Consider M sits first; 9! arrangements are possible.

For both W and C; they have 10 position and 10 people; so thats 10! each.

Also MWC can be in any order among themselves in a group; so thats 3! ways among themselves.

In total; 9! 10! 10! 3!

P.S In your solution; after all the women are seated, you have only 10 positions left, where does 20P10 for the children come from when women are already seated? Again I am unsure of my answer though. You got an OA?

I think 3! should not be there. else you may not be able to seat MWC alternatively in some of the cases. Here we have to maintain alternativity and so we dont have to change the order of any MWC set.

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by gmatclubmember » Fri Sep 23, 2011 5:57 am
Brent@GMATPrepNow wrote:
knight247 wrote:In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?

Here is my take. Lets begin with any group first eg Men. 10 Men can be seated around the table in 9! ways. 10 vacant seats are now left between the men which can be filled by the 10 women in 10! ways. 20 vacant spots are now left around the table and can be filled by the 10 children in 20P10 ways=20!/10!

So total number of ways=9!10!20!/10!=[spoiler]9!20![/spoiler] Is my approach correct?
The question seems pretty ambiguous to me. It's hard to tell whether the physical location (vs the relative location) of each seated person creates a new arrangement.
For example, if we have one arrangement, and then everyone shifts 1 seat to the left, does this create a new arrangement, even though the relative location of each person is the same?

Cheers,
Brent
Well I guess the chairs are not marked so all the combos are treated as ONE till the time the relative positions are same :).

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by knight247 » Fri Sep 23, 2011 6:18 am
@Brent
If the relative location of each person stays the same then its NOT a new arrangement. So we could have MCW MCW MCW MCW MCW etc which would be one arrangment or MWC MWC MWC which would be a different arrangement. Yeah, I know this problem is a real pain. Hope you can sort this out for us though. I unfortunately don't have an OA. Thanks Brent.

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by Brent@GMATPrepNow » Fri Sep 23, 2011 7:21 am
knight247 wrote:In how many different ways can you seat 10 women, 10 men, and 10 children around a circular table if you need to alternate the men, women and children?
I'm pretty sure this question is beyond the scope of the GMAT, but here's my attempt.

First, let's pretend that the location of the seats does matter. This means that, if we have one arrangement, and then everyone shifts 1 seat to the left, this creates a new arrangement, even though the relative location of each person is the same.

What are the implications here? We will be counting each arrangement more than once. In fact, we will be counting each arrangement 30 times, rather than just 1 time.

Why?

Well, if we take one arrangement, and then have everyone shift one seat to the left to get a "different" arrangement, we can keep doing this until each person has occupied all 30 seat.

So, if we assume that each that the location of the seat does matter, then the number of arrangements we calculate must later be divided by 30 to eliminate all of the duplication.

Okay, at this point, we can number the chairs #1, #2, #3 and so on.

We have 2 cases to consider:
case 1) the order is MWC MWC MWC etc
case 2) the order is MCW MCW MCW etc

Case 1
Let's place a man is seat #1. This means that the men will occupy seats, #1, 4, 7, 10,...
We can seat a man in seat #1 in 10 ways, then seat a man in seat #2 in 9 ways, etc.
So, we can seat the men in 10! ways.

Next we seat the women. They will occupy seats #2, 5, 8, 11, ...
We can seat a woman in seat #2 in 10 ways, then seat a woman in seat #5 in 9 ways, etc.
So, we can seat the woman in 10! ways.

Next we seat the children. They will occupy seats #3, 6, 9, 12,...
We can seat a child in seat #3 in 10 ways, then seat a child in seat #6 in 9 ways, etc.
So, we can seat the children in 10! ways.

So, we can seat all 30 people in (10!)(10!)(10!) ways [assuming that the location of each seat matters]

Important: Before we examine case 2, we should also recognize that I could have also seated the men in seats #2, 5, 8, 11,..., which means the women would have been seated in seats #3, 6, 9, 12,... and the children in seats #4, 7, 10, ...

So, if we seat the men in seats #2, 5, 8, 11,... (and then seat the women and children), we can accomplish this in (10!)(10!)(10!) ways [by applying the same logic as earlier]

Also, we can seat the men in seats #3, 6, 9, 12,..., which means the women would have been seated in seats #4, 7, 10, ... and the children in seats #5, 8, 11, ...
This can be accomplished in (10!)(10!)(10!) ways as well.

At this point, we have considered all of the possible MWC configurations.

So, for case 1, the total number of ways to seat all 30 people = (10!)(10!)(10!)+(10!)(10!)(10!)+(10!)(10!)(10!) = (3)(10!)(10!)(10!)


Case 2
Using the logic we used for case 1, we can seat all 30 people in (3)(10!)(10!)(10!) ways


Combining both case, we see that we can seat everyone in (6)(10!)(10!)(10!) ways [assuming that the location of each seat matters]

At this point, we can account for duplication by dividing by 30 to get:

(6)(10!)(10!)(10!)/30, which equals (10!)(10!)(10!)/5 ways

Proviso: I'm not feeling 100% about this solution - perhaps this morning's coffee hasn't kicked in yet :-)

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by gmatboost » Fri Sep 23, 2011 10:18 pm
The number of complex and ambiguously-worded combinations questions on BTG is amazing. You'd think that combinations was the number 1 topic on the GMAT Math section. The truth is, you'd be lucky to see more than one question on this topic.

It's especially unfortunate when people dedicate a lot of time to trying to solve questions that are ambiguous, because that is just never a problem on the GMAT. I do not buy into the idea that trying to parse this question will somehow make you better on easier combinations questions. It will just distract you from spending time on other concepts that are tested more often.

I absolutely agree with Brent:
[spoiler]I'm pretty sure this question is beyond the scope of the GMAT[/spoiler]

I also completely agree with his explanation.

The important thing is to understand the concepts that might actually show up again, not to worry about the wording or intended meaning.

Important takeaways from this question, in my opinion:
1. There are 6 ways to start out. MCW, MWC, CMW, CWM, WCM, MCW.
2. If arrangements that are relatively equivalent are considered different, there will be 30 times as many possibilities as if they are considered the same. (For example, is a circular table in which people are seated 1234 different from 2341, or is it the same?)
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by quantskillsgmat » Sat Sep 24, 2011 2:19 am
this question seems incomplete.however i assumes that group of each m ,w and children can be arranged.
10 sets of persons
so in circle they can be arranged in 9! ways.
further group can be start with men or women or children.
so 3 cases are possible.
men themselves can be arranged in 10! ways samw with womwn and children
so funal answer is
so my answer is 9!x3x10!x10!x10!

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by shankar.ashwin » Sat Sep 24, 2011 4:23 am
Well, I think circular permutation (n-1)! works only when there is only 1 element present. Here since there are 3 entities, treat the problem no different than a normal permutation problem.

In basic arrangements,

_ _ _ (Considering the first set of 3 to be seated)

Any of the 30 can seat in the first (M, W or C)

The second slot can be filled by 20 possibilites (Omitting M,W or C who have already seated in the first)

The third can be seated by any of the remaining 10 (again excluding 2 of M,W or C who have already been seated)

So, we have 30*20*10 for the first 3.

Here you have already accounted for order, so all of the remaining 9 M,W or C can fill up 9 slots in 9!*9!*9! ways..

Combining we have 6000* 9!*9!*9! ways. (which is (6)(10!)(10!)(10!) )

I think this is pretty straightforward. (Circular arrangements between 3 different elements does not make sense)