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.
Circular Permutations Theory Question
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Certainly something to think about. I'll mulsch it over in my head on my drive home! I like it. An interesting question.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.
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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
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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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
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
Vivian Kerr
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Thank you for all the "thanks" and "follows"!
GMAT Rockstar, Tutor
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https://www.yelp.com/biz/gmat-rockstar-los-angeles
Former Kaplan and Grockit instructor, freelance GMAT content creator, now offering affordable, effective, Skype-tutoring for the GMAT at $150/hr. Contact: [email protected]
Thank you for all the "thanks" and "follows"!
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A very neat solution!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