probability and combinatorics mixed question A bus follows its route through nine stations, and contains six passengers. What is the probability that no two passengers will get off at the same station? 
no detailed solution is required here but an idea of the general line of thought could be nice...

The problem cannot be solved, even approximately, unless we make some quite unreasonable assumptions. You are presumably expected to make these assumptions.
First, we will assume that for any passenger $P$, the passenger is equally likely to get off at any one of the $9$ stations.  Experience in bus riding will show you how very unreasonable that assumption is.
Second, we will assume that the choices the various passengers make are independent. This is ordinarily false: often people travel in groups of size $\ge 2$. 
But let us hold our noses and go on. Call the passengers A, B, C, D, E, F. 
Whatever choice A makes, the probability that B chooses a different stop is $\frac{8}{9}$.
Given that A and B have chosen different stops, the probability that C chooses a stop different from those of A and B is $\frac{7}{9}$.
Given that A, B, C have chosen different stops, the probability D chooses a stop different from the one they chose is $\frac{6}{9}$.
Continue. We conclude that (under our assumptions) the probability all $6$ get off at different stops is $\frac{8}{9}\cdot\frac{7}{9}\cdot \frac{6}{9}\cdot\frac{5}{9}\cdot\frac{4}{9}$. 
Another way: The bus driver goes to the various passengers, in alphabetic order, and asks them to tell her their stops as a number, $1$ to $9$. She then makes a "word" by writing down thse $6$ choices as a $6$-digit number. 
There are $9^6$ such numbers, all equally likely.
There are $(9)(8)(7)(6)(5)(4)$ such numbers with all digits distinct. For the probability, divide this by $9^6$. 
