TITLE: Combinatorics: Using Inclusion-Exclusion Principle
QUESTION [8 upvotes]: I'm not sure how to start this. I've been told to use inclusion-exclusion, but I don't know which properties to use.
Intuitively, I want to solve the problem like this:
$E_1$ = 1 person leaves stop 1, 2 people leave stop 1, 3 people leave stop 1,$\cdots$,5 people leave stop 1 (need to have at least 5 left for the rest of the stops)
$$\binom {10} 1 + \binom {10} 2+ \cdots+ \binom {10} 5$$
$E_2$ = if 1 person left $E_1$, then 1 person leaves stop 2, 2 people leave stop 2,$\cdots$,4 people leave stop 2
$$\binom 9  1 +\cdots + \binom 9 4$$
if 2 people left $E_2$, then 1 person leaves stop 2, 2 people leave stop 2,$\cdots$,3 people leave stop 2
$$\binom 8 1 + \binom 8 2 + \binom 8 3$$
...etc.
But I'm not sure how to link up each event from the last event so that the conditions are satisfied, which leads me to think that I am terribly wrong.

REPLY [2 votes]: Alternatively you could use Stirling numbers of the second kind.
This problem could be modeled as dividing $n$ distinct objects into $r$ distinct groups (at least one in each group) is given by $$r! \times \left\{\begin{matrix} n \\ r \end{matrix}\right\}$$
So we get $ 6! \times  \left\{\begin{matrix} 10 \\ 6 \end{matrix}\right\} = 16435440$