# Math Help - Permutations in a circle

1. ## 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

2. ## Re: Permutations in a circle

Hello, aonin!

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: . $2! \times 9!$

3. ## 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?

4. ## 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.

5. ## Re: Permutations in a circle

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

6. ## Re: Permutations in a circle

Then it becomes 3!. How come 2! ?

7. ## Re: Permutations in a circle

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.

8. ## Re: Permutations in a circle

thank you. Understood the concept now.

9. ## Re: Permutations in a circle

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?

10. ## Re: Permutations in a circle

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.