1. ## Circular permutation

Can anyone check my following questions answer thanks!

Q. How many ways are there to seat six people around a "circular table", where seating are considered to be the same if they can be obtained from each other by rotating the table?

Ans, (n-1)! for circular table and therefore our answer is (6-1)! = 5! = 120 ways

Is my answer correct? Also, can anyone please explain how we got (n-1)! formula?

2. Originally Posted by robocop_911
Could someone please explain how the $(n-1)!$formula for circular arrangement around a table is derived.
Consider six people standing in a line.

• $P_1, P_2 , P_3 , P_4 , P_5 , P_6$

They can be arranged in $6!$ distinct ways.

Now suppose we arrange these people around a circle. The following permutation are not unique.

• $P_1, P_2 , P_3 , P_4 , P_5 , P_6$
• $P_2, P_3 , P_4 , P_5 , P_6 , P_1$

As one can become the other form rotating the table.

To solve this problem we keep one person fixed and arrange the other around the fixed person. so there will be $(6-1)!$ unique arrangements.

I hope this clarifies where the $(n-1)!$ comes from.

Side note:

In particular cases you will not consider clockwise and anticlockwise wise circular arrangements as being unique. For example, consider arranging beads in a necklace clockwise and anticlockwise arrangements are not consider unique and you can "flip" the necklace around to go form one to another. So in particular cases you may need to use $\frac{(n-1)!}{2}$

Bobak

3. ## its combination not permuatation ( my mistake )!

the sum is under topic of "Combination" not "permutation"!
my mistake!!!

Is your explanation same for combination too? Since, you have given explanation for permutation.

4. Originally Posted by robocop_911
the sum is under topic of "Combination" not "permutation"!
my mistake!!!

Is your explanation same for combination too? Since, you have given explanation for permutation.
This is a problem about permutations, a combination is a selection of a given number of elements from a larger number without regard to their arrangement. Permutations deal with how ordered sets can be arranged.

Bobak

5. ## Didn't understand how (n-1)! comes from!

Originally Posted by bobak
Consider six people standing in a line.
• $P_1, P_2 , P_3 , P_4 , P_5 , P_6$
They can be arranged in $6!$ distinct ways.

Now suppose we arrange these people around a circle. The following permutation are not unique.
• $P_1, P_2 , P_3 , P_4 , P_5 , P_6$
• $P_2, P_3 , P_4 , P_5 , P_6 , P_1$

As one can become the other form rotating the table.

To solve this problem to keep one person fixed and arrange the other around the fixed person. so there will be $(6-1)!$ unique arrangements.

I hope this clarifies where the $(n-1)!$ comes from.

Side note:

In particular cases you will not consider clockwise and anticlockwise wise circular arrangements as being unique. For example, consider arranging beads in a necklace clockwise and anticlockwise arrangements are not consider unique and you can "flip" the necklace around to go form one to another. So in particular cases you may need to use $\frac{(n-1)!}{2}$

Bobak
Hello Bobak,

I didn't quite understand how $(n-1)!$ comes from...
Why do we have to keep one person "fixed" in the table and move the rest around the "fixed" person?

6. Originally Posted by robocop_911
Hello Bobak,

I didn't quite understand how $(n-1)!$ comes from...
Why do we have to keep one person "fixed" in the table and move the rest around the "fixed" person?
If one person is fixed on the table with $n$ seats you have $n-1$ seats left that are no longer in a circular arrangements as the people on the end points do not meet, so the problem is reduded to how many arrangements of $n-1$ people can you have on a line.

Bobak

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# Problem of distinct and circle permutations

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