Thread: No. of ways to seat around a table when seats are numbered

Three single men, two single women and two families take their places at a round table. Each of the two families consists of two parents and one child. Find the number of possible seating arrangements if the seats are numbered and each child sits between their parents.

My answer: 7!2!2!=20160 but answer is 31680

Originally Posted by Punch

Three single men, two single women and two families take their places at a round table. Each of the two families consists of two parents and one child. Find the number of possible seating arrangements if the seats are numbered and each child sits between their parents.
My answer: 7!2!2!=20160 but answer is 31680

There are eleven seats. Place the youngest child at the table and his/her parents on either side in ways.
There are six places to seat the other child because its parents must be able to sit on each side: ways.
Then there are to seat the others.
That product gives the correct answer.

@Plato: is it preferred that someone post a separate thread for each problem when all are related?

I feel like we might be helping you more if we find a way to explain how to attack these problems rather than illustrating each answer. Is this for a class? Are there any problems you have done where you did get the right answer by yourself?

Originally Posted by Plato

There are eleven seats. Place the youngest child at the table and his/her parents on either side in ways.
There are six places to seat the other child because its parents must be able to sit on each side: ways.
Then there are to seat the others.
That product gives the correct answer.

Why is it not possible to treat a family as a cluster? Such that we treat a family as one so there are now 7 people. So 7! ways of arranging them. And the parents can rotate, since there are 2 families, 2!2!.

Therefore 7!2!2!. Although i understood your workings, I dont see how this is wrong

Originally Posted by Punch

Why is it not possible to treat a family as a cluster? Such that we treat a family as one so there are now 7 people. So 7! ways of arranging them. And the parents can rotate, since there are 2 families, 2!2!. Therefore 7!2!2!. Although i understood your workings, I dont see how this is wrong

There are several questions there.
1) I did treat the family as a cluster. But in this case the child must sit between the two parents that can be done in only two ways.

2) If this were a row of eleven seats then would be only nine places to seat the first child. Then there are a number cases for the second depending on the placement of the first.

3) But this is a circular arrangement with eleven seats. So it has eleven choices for the first child and six for second. Being circular makes the question easier that if it were linear.