An elevator starts with 10 people, how many ways can all the people... Cases for each floor? An elevator starts with 10 people on the first floor of an 8 story building and stops at each floor.  In how many ways can all the people get off the elevator?
The only way I can think to do this is a case by case for each floor but that seems extremely messy for examples: all 10 get out on first floor, then all 9, but then the ways the remaining 1 can get out on the the next 6 floors... etc. Is there a better way to do this?
 A: look up "stars and bars" - you are placing 10 balls in 7 boxes ( unless you really want people to be able to get out on the first floor, in which case you have 8 boxes )
Most likely the question intends that you think of "a way" as 6 people getting out on the 4th floor and 4 people getting out on the 5th floor and is not making the distinction about exactly which people are getting out on every floor.
If we go with 7 floors we will have 8 bars separating the floors, and we must arrange 10 stars representing people getting out. so the example above would be represented as ...
floor #  2 3     4       5   6 7 8
        | | | ****** | **** | | | |

You must keep the first and last bars in the first and last positions, but other than that, any arrangement of 10 stars and 6 bars corresponds to a possible scenario the number of arrangements is 
$$ \frac{16!}{10! 6!} = \binom{16}{6} $$
if you really wanted to let people out on the first floor the answer would be $\binom{17}{7}$
A: Any of the $10$ people can get off at any floor. So you have ten variables that can assume values from $1$ to $8$, therefore $8^{10}$ possibilities.
A: I've been thinking about this problem recreationally for a while. I took combinatorics a couple years ago but I derived the stars/bars formula I learned from just thinking about it.
10 people = 10 stars
6 floors = 5 bars
(10 + 6) choose 5 = 16 choose 5
I thought I had the solution this morning, but what about when you give each person a unique name? For simplicity, let names just be capital letters A through J (10 names).
I first thought it would be (16 choose 5) * 10!, with the 10! being the permutation of letters A through J, but then I realized that for example:
If we let | denote the bars (floor separators), then
A B | C D | E | ... etc is the same as
B A | D C | E | ...
I'm not sure how to not overcount. Anyone have any thoughts?
