Expert: Scott A Wilson - 11/2/2011

Question
QUESTION: 1)Why for circular permutation atleast 3 distinct objects are required.
2)Will any arrangment of two objects involve a circular arrangments
3)Please explain with example or proof what is the meaning of "The number of ways in which n objects can be arranged around a circular table if the arrangments are made with respect to the table is n!"
Scot please help with examples

ANSWER: To think of in geometry, it takes three non-liner points to define a circle.

If we only have two points, a circular arrangement can't be described.

If we have two objects around a table, there is only 1 way of arranging them.

If we have A, B, and C around a table, start from where A sits.
The only arrangements are ABC and ACB.

If we have A, B, C, and D around a table, start by sitting A somewhere.
B could then have 3 positions, only 2 would be left for C, and then only 1 for D.
This means we have 3*2*1 = 3! positions.

As it turns out, if we have n people to arrange around this table,
the number of ways it can be done is (n-1)!, for where the first one sits makes no difference.

If it matters where they are with relation to the table, then the first one does matter,
and that makes the number of choices n!.


---------- FOLLOW-UP ----------

QUESTION: Scot ,
Thanks a lot for your very simple explanation regarding the first two questions.
It would of great help to me if you elaborate the last one with examples or simple diagram As I am unable to get it clearly.ie.
How The number of ways in which n objects can be arranged around a circular table if the arrangments are made with respect to the table is n!"

Answer
Lets call the people A, B, C, D, E, F, ...

Start with person A and ask him to sit down.
Since we don't care about where he sits, there is only 1 way for this to be done.
So for n = 1, there is only 1 way.

Invite person B to come in and have a seat as well.
Note that B can be near A or opposite A, but anywhere he sits,
they are still in the same order around the table, so there is
still only one way to do this, so have A sit across from B,
so they are equally space.

Invite person C to come in and have a seat as well.
Now A and B are sitting down already, and C can choose to sit on either side of them.
To give the order, starting with A, gives ABC or ACB, so there are 2 ways.
Whatever way they are sitting, ask them to adjust themselves to form a triangle.

Invite person D to come in and have a seat somewhere at the table where three people are sitting already.  If there order was ABC, the new order would be ADBC, ABCD, or ABCD.  If the order was ACB, the new order would be ADCB, ACDB, or ACBD.  That makes a total of 6 ways in which to sit.

So far we have, for x=number of people and y=number of orders, (1,1), (2,1), (3,2), (4,6).

Suppose the 1st four people were in the order ABCD.
Having the 5th person, E, choose a spot would give us AEBCD, ABECD, ABCED, or ABCDE.
Note that since they are in a circle, EABCD ir the same as ABCDE.
This is 4 choices, and the same applies to any of the 6 orders the 4 people could be in.
This means there are 4*6 = 24 ways they could be seated.

We now have (1,1), (2,1), (3,2), (4,6), (5,24).
The series of y values is 1, 1, 2, 6, 24, and this is (y-1)!

Since y is really the same as n, this is (n-1)!, which is the same as n!/n.