Here is a way of visualizing the situation, in your case of $n=5$.
There are $5$ chairs, evenly spaced around a circular table. These chairs are labelled $1,2,3,4,5$, say counterclockwise.
We want to seat $5$ people $A,B,C,D,E$, so that seating arrangements that differ by a rotation are considered the same. The string $BAEDC$ means $B$ is at $1$, $A$ at $2$, $E$ at $3$, $D$ at $4$, and $C$ at $5$. This is considered the same as $AEDCB$, also $EDCBA$, $DCBAE$, and $CBAEDE$.
Call these $5$ arrangements a family. Any arrangement belongs to a unique family of $5$.
There are $5!$ arrangements of the letters $A,B,C, D, E$ in a line. We ask how many families there are. Well, since each family contains $5$ strings, there are $\dfrac{5!}{5}=4!$ families. Thus there are $4!$ circular permutations of our $5$ people.
With $n$ people, exactly the same idea works, and there are $\dfrac{n!}{n}=(n-1)!$ circular permutations.
Here we considered two arrangements, one of which is clockwise, and the other anti-clockwise, as different.
But if we consider two such arrangements as being the same, then for example $ABCDE$ and $EDCBA$ are considered the same. This means (in our case) that now each family has $10$ objects, twice as many as before. So the number of "really different" arrangements in this case is $\dfrac{5!}{(2)(5)}$.
This generalizes nicely to $n$ people, with a couple of minor exceptions.
If $n\ge 3$, the number of circular arrangements with clockwise and anti-clockwise considered the same is $\dfrac{n!}{2n}=\dfrac{(n-1)!}{2}$.
If $n=1$, obviously there is only $1$ arrangement, whatever the conditions. And if $n=2$, the one and only arrangement is the same, whether clockwise or anticlockwise are considered the same or not.
Best Answer
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 $$\frac{ 11\cdot 12!\cdot2}{14!}=\frac{11}{7\cdot 13}=\frac{11}{91}\,.$$