Seating arrangement probabilites Suppose that n people are seated in a random manner in a row of n theater seats. What is the probability that 2 particular people A and B will be seated next to each other?
So I think that the number of possible combinations would be : $n^n$ but I am not sure about the second part, with A and B sitting next to each other. I need some help about how to get started thinking about this problem
 A: Sasha has given the standard answer.  Here is a different approach using the fact that each arrangement is equally likely, but which avoids factorials or conjoined people. 
Seat A first.  Then the probability that B is seated immediately on A's right is $\dfrac1{n-1}$ (since there are $n-1$ people who are not A) multiplied by the probability there is in fact a seat to A's right $\dfrac{n-1}{n}$ (since if A sits on the far right there is no seat to the right), which is $\dfrac{1}{n}.$  
Similarly the probability that B is seated immediately on A's left is $\dfrac1{n}$.  So the probability they are sitting together is  $\dfrac2{n}$.
If the people were sitting at a round table, it should be obvious the answer would be $\frac2{n-1}$ when $n \gt 2$.
A: Let's count the total number of possible arrangements. The first person has choice of $n$ seats, the second of $n-1$, the next of $n-2$ and so on. The total number of possible seatings thus equals:
$$  
   N_T = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1 = n!
$$
Since $A$ and $B$ must sit together, we replace the pair $(A,B)$ with a meta person, hence reducing the number of (meta-) people to $n-1$. There are $N_{(A,B)} = (n-1)!$ ways of seating these. 
The probability is the ratio of the number of configuration of interests to the total number of possible configurations, under the assumptions that each configuration is equally likely:
$$
   \Pr\left(A\text{ seats next to }B\right) = \frac{N_{(A,B)} + N_{(B,A)}}{N_T} = \frac{2 (n-1)!}{n!} = \frac{2}{n}
$$
