..how many seating arrangements are possible? A Middle East peace conference will be attended by 5 Arab countries, 4 western countries, and Israel.  Each delegation will sit together at a round table subject only to the requirement that no Arab delegation be seated next to the Israeli delegation.  How many seating arrangements are possible?
 A: The trick is to choose two western countries to "buffer" the Israeli delegation. There are $\binom42\times 2=12$ ways to do this, accounting for order as well as the selection of $2$ out of $4$. Once this block is determined, the remaining $7$ delegations can be placed in the remaining seats in $7!$ possible ways. Thus, the answer is:
$$\binom42\times 2\times 7! = 6\times 2\times 5040 = 60480$$
A: Let $S$ be the set of all possible seating arrangements for the round table. Don't let the table's circular shape fool you. Unlike the typical case in which people are lined up, a circular table has neither a defined beginning nor a defined end. For example, the seating arrangement $\{\text{a},\text{b},\text{c},...,\text{j}\}$ is identical to $\{\text{b},\text{c},...,\text{j},\text{a}\}$, because for circular arrangements, you can arbitrarily select where you start indexing. A simple trick is to fix an arbitrary person to a seat, and then count the ways in which the remaining people can be seated. Here, this gives $|S| = (10-1)! = 9!$.
Now, we can deal with the constraint: No Arab delegate can sit beside the Israeli delegate. Let $C \subset S$ denote the subset of seating arrangements in which at least one Arab Delegate sits next to the Israeli delegate. As mentioned before, for circular tables, we have to fix a person. Intuitively, it seems like fixing the Israeli delegate will simplify the calculations, so let's go with that.
Fix the Israeli delegate to a seat, and there are $\frac{5!}{(5-2)!} = \frac{5!}{3!}$ ways in which two Arab delegates can be placed next to him, since we're counting the number of ways in which 2 people can be selected from 5 people with no replacement, and with ordering considered. Additionally, for each such arrangement, there are 7 remaining seats whose occupants can be re-arranged in $7!$ ways. So we have $\frac{7!5!}{3!}$ ways in which 2 Arab delegates can sit beside the Israeli delegate.
We still have to consider seating arrangements in which exactly one Arab delegate sits beside the Israeli delegate. Simple enough: There are 5 Arab delegates available to place at his left, just as there are 5 Arab delegates to place at his right. Therefore, we have 10 ways to place one Arab delegate beside the Israeli delegate. For each such case, there are 4 western delegates to place on the opposite side of the Israeli delegate, and $7!$ ways in which the remaining people can be re-arranged: A total of $(40)(7!)$ seating arrangements in which one Arab delegate sits beside the Israeli delegate. Combining everything, we get the answer
$$|S| -|C| = 9! - \frac{7!5!}{3!} - (40)(7!) = 60480.$$
Note that the portion of seats which are viable is exactly $\frac{1}{6}$:
$$\frac{|S|-|C|}{|S|} = \frac{9! - \frac{7!5!}{3!} - (40)(7!)}{9!} = \frac{1}{6}.$$
First and foremost, the fact that the answer is so clean is reassuring. Secondly, since Arab delegates form half of the group, we're counting the arrangements in which the Israeli delegate is completely surrounded by a minority group. So the fact that these seats represent a minority of the total seats available makes intuitive sense to me. 
