A problem of circular permutation

[sahilmm15]

I have understood everything till 2nd paragraph. But after that everything was above my head. Can you explain me the concept

 

[PeroK]

What in particular isn't clear?

 

[sahilmm15]

What in particular isn't clear?

I am not clear on how they derived the formula at last.

 

[PeroK]

I am not clear on how they derived the formula at last.

You put ABCD in a line, there are $4! = 24$ ways to do it.

You put them in a circle, there are $4!/4 = 6$ ways to do it.

This assumes that it's only the order round the table that is important (who is sitting next to whom). If the precise positions are important, then it's still $4!$ for the circle.

 

[sahilmm15]

You put ABCD in a line, there are $4! = 24$ ways to do it.

You put them in a circle, there are $4!/4 = 6$ ways to do it.

This assumes that it's only the order round the table that is important (who is sitting next to whom). If the precise positions are important, then it's still $4!$ for the circle.

Thanks for the reply. It's clear now.

 

[sahilmm15]

You put them in a circle, there are $4!/4 = 6$ ways to do it.

By the way why you divided $4!$ by 4. I founded that the order doesn't matter for the first person, there would be only 1 way to sit. After the first person sits, now the order matters, because each arrangement would not be the same unlike in the first case where arrangement was same every time. So $1*3*2=6$

 

[PeroK]

Let's do it for three places. For a row we have:

ABC, ACB, BAC, BCA, CAB, CBA

That's $3! = 6$ ways.

For a circle, if we label the places then there are also six ways, exactly as above. But, if we consider only the order round the table, then:

ABC, CAB, BCA are all the same; and ACB, BAC, CBA are all the same. So, we have only two ways.

 

[sahilmm15]

Let's do it for three places. For a row we have:

ABC, ACB, BAC, BCA, CAB, CBA

That's $3! = 6$ ways.

For a circle, if we label the places then there are also six ways, exactly as above. But, if we consider only the order round the table, then:

ABC, CAB, BCA are all the same; and ACB, BAC, CBA are all the same. So, we have only two ways.

So, the observation of the above fact led people to generalize it with the formula. That's cool. Now I understood why you divided$4!$ by $4$

 

[PeroK]

So, the observation of the above fact led people to generalize it with the formula. That's cool. Now I understood why you divided$4!$ by $4$

Yes, and if you have $n$ people round the table, then the same arrangement (in terms of who is sitting next to whom) can be achieved in $n$ ways. You start with an arrangement and then you rotate the whole table by an nth of a rotation and it's still the same.