Problem: 15 people, each from different countries, sit around a circular table. What is the probability that a British and a French representative sits next to each other but an American and a Russian representative doesn't.
The correct solution for this problem is $\frac{11}{91}$.
I know that all the different variations are $\frac{15!}{15} = 87\ 178\ 291\ 200$, but how can I count all the arrangements that satisfy the given conditions?
On
There are $14!$ ways to sit the $15$ people with no constraints.
There are $2\cdot 13!$ to seat the table with the Brit and the Frenchman next to each other.
There are $4\cdot 12!$ to seat the table with the Brit and the Frenchman next to each other and the American and the Russian next to each other.
$\frac {2\cdot 13! - \frac 4\cdot 12!}{14!}\\ \frac {2\cdot 13 - 4}{14\cdot 13}\\ \frac {11}{91}$
Treat the English and the French representatives as a single individual. Basically, we are seating $14$ individuals in a way that the American and the Russian do not get close to one another, lest they fight. Fix a seat for the American. Then, there are $11$ possible seats for the Russian. Now, for the remaining $12$ individuals, there are $12!$ ways to arrange them. Finally, we de-cluster the English and the French, and there are $2!=2$ ways to permute them. Hence, there are in total $$11\cdot 12!\cdot 2$$ ways to arrange the $15$ people according to the rules.
Now, there are $(15-1)!=14!$ arrangements without care to the seating rules. Thus, the probability that a random seat assignment will be satisfactory is $$\frac{ 11\cdot 12!\cdot2}{14!}=\frac{11}{7\cdot 13}=\frac{11}{91}\,.$$