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Circular Permutations Theory Question

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Circular Permutations Theory Question

by NickPentz » Wed Nov 13, 2013 10:34 am
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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by Mathsbuddy » Wed Nov 20, 2013 9:30 am
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.
Certainly something to think about. I'll mulsch it over in my head on my drive home! I like it. An interesting question.
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by Mathsbuddy » Wed Nov 20, 2013 9:59 am
7 people, 5 seats,
I would look at the probability of one particular person sitting in each seat:

Seat 1: 1/7
Seat 2: 1/6
Seat 3: 1/5
Seat 4: 1/4
Seat 5: 1/3

P(pre-selected arrangement) = Product of all above = 1/2520

Hence there are 2520 possible arrangements.

2520/5 = 504 unique arrangements in the circle.

I think 1*6*5*4*3= 360 is not right because the first person you "fix" is from a choice of 7 people.
It means that Mr Fixed is sitting is guaranteed a seat in every arrangement. This model misses out the arrangements when he is not at the table.

Consider a straight line example:

People pool: FGHIJKL

If F is Mr Fixed, then he will be sitting in any of these seats:
F~~~~ * 360 arrangements of the others
~F~~~ * 360 arrangements of the others
~~F~~ * 360 arrangements of the others
~~~F~ * 360 arrangements of the others
~~~~F * 360 arrangements of the others

But what about other times when he is not there?
~~~~~ * 6 * 5 * 4 * 3 = 360 when he is sitting outside the room in seat 6
~~~~~ * 6 * 5 * 4 * 3 = 360 when he is sitting outside the room in seat 7

360*7 = 2520

In a circle divide by 5 to get 504 again.

Hence 504 is correct.

I hope this helps :)
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by VivianKerr » Fri Dec 06, 2013 12:23 am
It's like a Combination AND a Permutation - I like it!

So first we're wondering, how many ways to choose 5 from 7? This is a simple Comb:

7C5 = 7! / 5!2! = 7 x 6 / 2 = 42/2 = 21 ways

So now for each of those ways, we're wondering, how many ways can we order 5 people around a table?

For any table with "x" seats, the number of possible arrangements is (x-1)!, so here 4! = 4 x 3 x 2 = 24.

21 x 24 = 504
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by Mathsbuddy » Fri Dec 06, 2013 1:01 am
VivianKerr wrote:It's like a Combination AND a Permutation - I like it!

So first we're wondering, how many ways to choose 5 from 7? This is a simple Comb:

7C5 = 7! / 5!2! = 7 x 6 / 2 = 42/2 = 21 ways

So now for each of those ways, we're wondering, how many ways can we order 5 people around a table?

For any table with "x" seats, the number of possible arrangements is (x-1)!, so here 4! = 4 x 3 x 2 = 24.

21 x 24 = 504
A very neat solution!
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