# arrangement in circle problem

• Apr 21st 2010, 03:44 AM
lemons
arrangement in circle problem
A problem in my textbook asks:
1. In how many ways can four ours and four girls be arranged in a circle? In how many of these will boys and girls occur alternately?
2. In how many ways can n boys and n girls be arranged in a circle? In how many of these will the boys and girls occur alternately.

The first part of both questions is easy, 7! and (2n-1)!
But I'm really confused on the second parts.

I worked it out as the first boy to sit begins the circle and everyone is placed relative to them. Then the next boy sits can sit to the left, or the right, or opposite the first boy. Then the third has two choices, and the fourth one. And then the girls sit down, the first has a choice of four places to sit, the second three, etc

So I get the answer: (4-1)! * 4! = 144
and for n boys and n girls: (n-1)! n!

but my book says the answer should be 72
and it says for n girls and n boys, the answer is 0.5 n!(n - 1)!
but this doesn't make sense to me, especially as if you only have one boy and one girl then their answer would be half an arrangement, which is nonsense?
• Apr 21st 2010, 06:16 AM
undefined
Quote:

Originally Posted by lemons
A problem in my textbook asks:
1. In how many ways can four ours and four girls be arranged in a circle? In how many of these will boys and girls occur alternately?
2. In how many ways can n boys and n girls be arranged in a circle? In how many of these will the boys and girls occur alternately.

The first part of both questions is easy, 7! and (2n-1)!
But I'm really confused on the second parts.

I worked it out as the first boy to sit begins the circle and everyone is placed relative to them. Then the next boy sits can sit to the left, or the right, or opposite the first boy. Then the third has two choices, and the fourth one. And then the girls sit down, the first has a choice of four places to sit, the second three, etc

So I get the answer: (4-1)! * 4! = 144
and for n boys and n girls: (n-1)! n!

but my book says the answer should be 72
and it says for n girls and n boys, the answer is 0.5 n!(n - 1)!
but this doesn't make sense to me, especially as if you only have one boy and one girl then their answer would be half an arrangement, which is nonsense?

I get what you get. For example, if n = 2, it is clear to me that there are two arrangements, which is another counterexample to the answer given by the book. The only way I could imagine the book answer being right is if two arrangements are considered equivalent up to rotation and reflection, and n > 1, but those assumptions seem inappropriate to the situation.
• Apr 24th 2010, 01:29 AM
lemons
Thank you, that is exactly what I thought.
• Apr 24th 2010, 06:24 AM
Quote:

Originally Posted by lemons
A problem in my textbook asks:
1. In how many ways can four ours and four girls be arranged in a circle? In how many of these will boys and girls occur alternately?
2. In how many ways can n boys and n girls be arranged in a circle? In how many of these will the boys and girls occur alternately.

The first part of both questions is easy, 7! and (2n-1)!
But I'm really confused on the second parts.

I worked it out as the first boy to sit begins the circle and everyone is placed relative to them. Then the next boy sits can sit to the left, or the right, or opposite the first boy. Then the third has two choices, and the fourth one. And then the girls sit down, the first has a choice of four places to sit, the second three, etc

So I get the answer: (4-1)! * 4! = 144
and for n boys and n girls: (n-1)! n!

but my book says the answer should be 72
and it says for n girls and n boys, the answer is 0.5 n!(n - 1)!
but this doesn't make sense to me, especially as if you only have one boy and one girl then their answer would be half an arrangement, which is nonsense?

Hi lemons,

this is offered as an explanation for the book solution,
not as "the answer" as there will be disagreement.

A boy and girl can only form a line.
We need at least 3 to make a circle.

If we have 3, then we have ABC, ACB as arrangements,
as we can keep A still and move the other two,
but as far a making a circle is concerned it is always the circle ABC.
This is the view your book is taking.

Though they could change hands, if we draw a circle and label the 3 points
A, B, C with A at the top, B to the left and C to the right, then reading it anticlockwise it is ABC and if we interchange B and C,
then clockwise it is ABC, so it's considered the same circle.

As far as lining them up in a straight line is concerned,
there are twice as many arrangements with A fixed.

For 4 girls and 4 boys...

If we fix one girl in place at the beginning of the line, there are $3!4!$ straight line arrangements
of the remaining girls and boys alternating.

If they are arranged in a circle, that number is halved.....from the perspective of the circle being the same forwards or backwards

This would appear to be the view that your textbook is taking,
but that is only one view and according to a particular interpretation.

From another view, it will be considered "incorrect",
but only because the textbook hasn't fully clarified the question.
• Apr 24th 2010, 07:23 AM
Plato
Quote:

A boy and girl can only form a line.
We need at least 3 to make a circle.

If we have 3, then we have ABC, ACB as arrangements,
as we can keep A still and move the other two,
but as far a making a circle is concerned it is always the circle ABC.
This is the view your book is taking.

For 4 girls and 4 boys...
If we fix one girl in place at the beginning of the line, there are $3!4!$ straight line arrangements of the remaining girls and boys alternating.
If they are arranged in a circle, that number is halved.

The above reply has several issues. This is such a well know problem.
First, we can certainly seat one boy and one girl at a circular table in one way.
Given a circular table with four seats.
We can seat the two girls opposite one another and the there only two ways to seat the two boys.

Now consider the case of 4 girls, ABCD, and 4 boys, WXYZ.
Seat A any where at table. Then seat BCD in ever other seat beginning at A’s left.
This can be done in $3!$ ways. Then the boys can take the remaining seats in $4!$ ways.
Thus for the case $N=4$ the answer is $(3!)(4!)$.
Dividing by two is incorrect.