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?
Best Answer
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 ...
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 $\binom{17}{7}$