Circular Permutation

[Agent Smith]

So I have 5 people. I want to arrange them (permute them) in a circle.

My answer is [imath]\frac{^5P_5}{5 \times 2} = \frac{5!}{10} = 12[/imath]

Does that mean for r objects to be permuted in a circle from n objects, we do this [imath]\frac{n!}{2r}[/imath]?

 

[Agent Smith]

 

[Dr.Peterson]

So I have 5 people. I want to arrange them (permute them) in a circle.

My answer is [imath]\frac{^5P_5}{5 \times 2} = \frac{5!}{10} = 12[/imath]

Does that mean for r objects to be permuted in a circle from n objects, we do this [imath]\frac{n!}{2r}[/imath]?

Why would you think that a formula for arranging all of 5 people can be directly extended to one for some of r objects? You're going from one variable to two variables, so, no, it doesn't "mean" that. You need to apply similar thinking to the new problem.

Also, you're assuming your answer to the first question is correct. Can you explain why? And, since your answer is a small number, have you tried listing the permutations as a check on your thinking? (That could also show if you mean something different from what I would mean by the question, which is conceivable.)

 

[Agent Smith]

@Dr.Peterson It doesn't work for 2 people [imath]\frac{2!}{2\times 2} = \frac{1}{2}[/imath].

I think the formula works for 3 people [imath]\frac{3!}{3\times 2} = 1[/imath]

 

[khansaheb]

@Dr.Peterson It doesn't work for 2 people [imath]\frac{2!}{2\times 2} = \frac{1}{2}[/imath].

I think the formula works for 3 people [imath]\frac{3!}{3\times 2} = 1[/imath]

View attachment 38871

And the second arrangement can be B and C change position keeping position of A same. Any more - think again

 

[Agent Smith]

And the second arrangement can be B and C change position keeping position of A same. Any more - think again

I didn't realize that being on the left or right of an object mattered.

Should I multiply by 2 then?

We have [imath]\frac{n!}{n} = n-1[/imath] permutations?

For 4 objects we have my [imath]\frac{(4 - 1)!}{2} \times 2 = 6[/imath]

The formula works then for 2, [imath]\frac{1}{2} \times 2 = 1[/imath].

Please correct me, if wrong.

 

[Agent Smith]

I mean [imath]\frac{n!}{n \times 2} \times 2 = (n - 1)![/imath]. The story checks out, I guess.

 

[Dr.Peterson]

Try looking it up:

Circular Permutation -- from Wolfram MathWorld

The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is P_n=(n-1)!. The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated...

mathworld.wolfram.com

What you calculated at first was "free circular permutations"; now you've corrected it to "circular permutations", where the direction of the circle matters ("cannot be picked up out of the plane and turned over").

And, of course, none of this involves the "r objects from n objects" of your question.

 

[Agent Smith]

Try looking it up:

Circular Permutation -- from Wolfram MathWorld

The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is P_n=(n-1)!. The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated...

mathworld.wolfram.com

What you calculated at first was "free circular permutations"; now you've corrected it to "circular permutations", where the direction of the circle matters ("cannot be picked up out of the plane and turned over").

And, of course, none of this involves the "r objects from n objects" of your question.

If r = n, then?

Gracias for the link. Thanks for giving the names of the arrangements.