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Circular Permutations Question

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Circular Permutations Question [#permalink] New post 14 Nov 2013, 14:43
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Hello, I have not seen a question about circular permutations involving more x than y in the circle. Could someone explain how this works to help me fill in this gap?

For example, there are 7 people and a round table with 5 seats. How many arrangements are possible?

My guess would be to fix the first person and work from there:
1*6*5*4*3= 360
I can also think of starting with 7 and then dividing by the number of seats since each arrangement can be rotated 5 times while still being the same arrangement:
(7*6*5*4*3)/5= 504
Can someone explain which is right and why?

Next example, there are 4 people and 5 seats at a round table. How many arrangements are possible?

My guess here is to treat the empty seat just like another person:
(5-1)! = 4! = 24

Thanks for the help.
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Re: Circular Permutations Question [#permalink] New post 15 Nov 2013, 16:51
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NickPentz wrote:
Hello, I have not seen a question about circular permutations involving more x than y in the circle. Could someone explain how this works to help me fill in this gap?

For example, there are 7 people and a round table with 5 seats. How many arrangements are possible?

My guess would be to fix the first person and work from there:
1*6*5*4*3= 360
I can also think of starting with 7 and then dividing by the number of seats since each arrangement can be rotated 5 times while still being the same arrangement:
(7*6*5*4*3)/5= 504
Can someone explain which is right and why?

Next example, there are 4 people and 5 seats at a round table. How many arrangements are possible?

My guess here is to treat the empty seat just like another person:
(5-1)! = 4! = 24

Thanks for the help.

Dear Nick,
I'm happy to respond. :-)

The first method you gave, ...
1*6*5*4*3= 360
... doesn't work because for that first seat there is more than one choice. That would work if there were five people and five seats: then we could fix where Bertha is sitting, and arrange the other 4 people accordingly, (1)*(4!). Here, there's nothing "fixed' about that fixed seat, if you see what I mean.

I would say your second method,
(7*6*5*4*3)/5= 504
is golden. We use the Fundamental Counting Principle to count the total number of seatings, and then divide by 5 to eliminate circular redundancy. That's the way to do this.

On the second problem, yes, you can treat the empty seat as a person (perhaps a more interesting person than some of the others, who knows?), and the answer you calculated is correct. That's already probably harder than anything the GMAT would give you. The problem becomes a bit hairy if there is more than one empty seat --- for example, 7 seats and 4 people: that is beyond what the GMAT would expect anyone to know.

Great questions and great work!
Mike :-)
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Re: Circular Permutations Question [#permalink] New post 15 Nov 2013, 23:43
Nick,

Slightly different but essentially the same result for the first problem. First find out the number of ways to select a group of 5 people from a set of 7 people. This is 7C5 which is equal to 21. Now each one of these sets of 5 people can be arranged in (5-1)!=4!=24 ways around the circle. The total number of such arrangements is (21)(24)=504. Here I am directly using the result, without giving a conceptual explanation, that n objects can be arranged in (n-1)! ways around the circle.

So far on the GMAT, I have only seen one circular permutation problem, and would expect any new question to likely fall in the 750+ level. If you are not consistently scoring Q48+ in the quant, I would recommend holding off spending too much time on advance permutation and combination problems.

Cheers,
Dabral
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Re: Circular Permutations Question [#permalink] New post 17 Nov 2013, 20:00
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NickPentz wrote:
Hello, I have not seen a question about circular permutations involving more x than y in the circle. Could someone explain how this works to help me fill in this gap?

For example, there are 7 people and a round table with 5 seats. How many arrangements are possible?

My guess would be to fix the first person and work from there:
1*6*5*4*3= 360
I can also think of starting with 7 and then dividing by the number of seats since each arrangement can be rotated 5 times while still being the same arrangement:
(7*6*5*4*3)/5= 504
Can someone explain which is right and why?

Next example, there are 4 people and 5 seats at a round table. How many arrangements are possible?

My guess here is to treat the empty seat just like another person:
(5-1)! = 4! = 24

Thanks for the help.


Note that with 7 people and 5 seats, you can make only 5 people sit. So another way of going about it is you first select the 5 people who are going to sit in 7C5 ways. Now you simply have 5 people and 5 seats around a circle. You can make them sit in 4! ways (we know the (n-1)! formula)
So you get 7C5*(4!) = (7*6/2)*4! = 504
If number of people are more than the seats, the problem can simply be brought down to equal number of people and seats by selecting the people who will sit.

If the number of seats are more, you can increase the number of people by bringing in Mr V (a vacant spot). If you have more than one vacant spots, all Mr Vs are considered identical.
Check out the concept of Mr V (the vacant spot) in linear arrangements here: http://www.veritasprep.com/blog/2011/10 ... ts-part-i/

Or you can use the same basic counting principle to arrive at the answer in this case too.
Say you have 4 people and 5 seats around a circle. Every person will sit. So start with the first person. He has 1 way of sitting around a circular table because all seats are identical.
Second person has 4 options. (since all seats are distinct now)
Third person has 3 options.
Fourth person has 2 options.
So all 4 can sit in 1*4*3*2 = 24 ways

Now you can extend this logic with any number of seats.
Say you have 4 people and 8 seats around a circle. Again, every person must sit.
Start with the first person who has 1 way since all seats are identical.
The second person can sit in 7 ways since the rest of the 7 seats are distinct now.
The third person in 6 ways.
The fourth person in 5 ways.
So all 4 can sit in 1*7*6*5 = 210 ways.
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Re: Circular Permutations Question   [#permalink] 17 Nov 2013, 20:00
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