I have an issue with two problems with round table.
I have $n$ families, each family have father, mother, kid and a dog.
First problem: I need to sit the families that every kid sits between his parents.
Second problem: I need to sit the families that the two parents sit together and no dogs near each other.
In the first problem I tried to stick the parents and the kid together like this:
I have total of $2n$ places in the table, so to sit the parents and kid between them I have $n!$ options (and multiplies by $2$ because the side the parents sit) and $(n-1)!$ for the dogs to sit.
$$n!\cdot 2\cdot (n-1)!$$
but I'm not really sure about it.
And in the second problem I thought to stick the parents together and the kid and the dog together, but it seems to be wrong because I need the dog the be alone because there can be a possibility that the dog can be between two couples of parents.
Thank you very much for your help!
Best Answer
For the first problem, the factor $2$ applies to each family, so that would give $2^n$. Presumably the dogs need to go between the families, but dogs can be together so you are ordering $2n$ (families plus dogs). The result is $2^n(2n-1)!$
For the second, you have $n$ couples, $n$ kids, and $n$ dogs to seat. You again get a factor $2^n$ for the couples' order. Then attach a couple or kid to the right of each dog. You have $2n$ choices for the first dog, $2n-1$ for the second, and so on. This gives a factor $\frac {(2n)!}{n!}$. Now you have $2n$ groups to arrange around the circle, which you can do in $(2n-1)!$ ways, for a final result of $2^n\frac {(2n)!(2n-1)!}{n!}$