People around a circular table problem

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?

• Seat the British and French representatives next to one another. Then the problem is reduced to the probability that the American and Russian don't sit next to one another in a line of $13$ people. So, you have two events whose probabilities you need to work out. – saulspatz Nov 7 '18 at 21:08
• So if I seat first British and French next to one another, I have $15*2$ alternatives. Then I seat an American and a Russian so that they don't sit next to each other in a line of $13$ people, I have $(13*10+2)*11!$ alternatives. Combined I have $15*2*(13*10+2)*11!$ alternatives. I don't get the right result. What I am missing here? – jte Nov 7 '18 at 21:20
• The probability that the British and French ambassadors sit next to one another is $1/7$ isn't it? Wherever the British ambassador sits, there are $2$ out of $14$ places the French ambassador can sit. – saulspatz Nov 7 '18 at 21:26

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 $$\frac{ 11\cdot 12!\cdot2}{14!}=\frac{11}{7\cdot 13}=\frac{11}{91}\,.$$
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.
$$\frac {2\cdot 13! - \frac 4\cdot 12!}{14!}\\ \frac {2\cdot 13 - 4}{14\cdot 13}\\ \frac {11}{91}$$