# Permutations in a circle

• Sep 7th 2011, 04:59 PM
aonin
Permutations in a circle
Q:Three engineers and nine coucillors have a meeting around a circular table. If three coucillors are between each engineer, find the number of possible seating arrangments.

the circle permutations formula is $(n-1)!$

the answer is Arrangements: $2! \times 9!$

whats wrong with $3!\times8!$, could someone explain why this is wrong, why is it not adhering to the conditions of the question?

Thanks
• Sep 7th 2011, 07:32 PM
Soroban
Re: Permutations in a circle
Hello, aonin!

Quote:

Three engineers and nine coucillors have a meeting around a circular table.
If three coucillors are between each engineer,
. . find the number of possible seating arrangments.

The answer is: . $2! \times 9!$

The basic arrangement looks like this (not including rotations):

. . $\begin{array}{c} E \\ C \;\;C \\ C\;\;\;\;\;\;C \\ C \;\;\;\;\;\;\;\;\;\;C \\ E \;\;C\;\;C\;\;C\;\;E \end{array}$

The first engineer can sit in any chair (it doesn't matter).
The second engineer has a choice of either of the other two E-chairs.
The third engineer has only one choice: the remaining E-chair.
. . Hence, the engineers can be seated in $2!$ ways.

Then the nine "coucillors" can be seated in $9!$ ways.

.Answer: . $2! \times 9!$

• Sep 7th 2011, 09:48 PM
aonin
Re: Permutations in a circle
Thanks soroban, but why are the engineers seated down first? Is it possible to sit the 9 councillors down then only the engineers?
• Sep 7th 2011, 11:44 PM
arangu1508
Re: Permutations in a circle
Dear Soroban,

When the first engineer can sit in any chair, means he has 12 chairs in his choice is it not? or at least 3 chairs in his choice. Kindly explain me.
• Sep 8th 2011, 01:48 AM
aonin
Re: Permutations in a circle
Quote:

Originally Posted by arangu1508
Dear Soroban,

When the first engineer can sit in any chair, means he has 12 chairs in his choice is it not? or at least 3 chairs in his choice. Kindly explain me.

The first engineer can sit in any 3 chairs, the next engineer 2 chairs, the last engineer can only sit in 1 chair
• Sep 8th 2011, 10:19 AM
arangu1508
Re: Permutations in a circle
Then it becomes 3!. How come 2! ?
• Sep 8th 2011, 10:35 AM
Plato
Re: Permutations in a circle
Quote:

Originally Posted by arangu1508
Then it becomes 3!. How come 2! ?

Consider an empty table. Put three chairs there.
Now seat the three engineers. Using circular permutations* that can be done in $(3-1)!=2!$ ways.
At this point the table is ordered. Place three chairs between each of the engineers. Now there $9!$ ways to seat the councilors.

*circular permutations: there are $(n-1)!$ ways to arrange n different objects in a circle.
• Sep 8th 2011, 11:09 PM
arangu1508
Re: Permutations in a circle
thank you. Understood the concept now.(Giggle)
• Sep 9th 2011, 03:47 AM
aonin
Re: Permutations in a circle
Quote:

Originally Posted by Plato
Consider an empty table. Put three chairs there.
Now seat the three engineers. Using circular permutations* that can be done in $(3-1)!=2!$ ways.
At this point the table is ordered. Place three chairs between each of the engineers. Now there $9!$ ways to seat the councilors.

*circular permutations: there are $(n-1)!$ ways to arrange n different objects in a circle.

this might be counterintuitive, but what happens if you decide to seat the councillors first? then the engineers?
• Sep 9th 2011, 04:48 AM
Plato
Re: Permutations in a circle
Quote:

Originally Posted by aonin
this might be counterintuitive, but what happens if you decide to seat the councillors first? then the engineers?

One could do that. But it makes the explanation more difficult.
$2(9!)=9(2(8!))=3((3!)8!)$
With three at the table, there is only one way to add nine more chairs in the required way.
But with nine already at the table how ways are there to add three more chairs in the required way?
Each engineer is between two councilors in the first way,
But the same cannot be said of the councilors.
We look for the least complicated way of modeling.