In how many ways can people get out of the elevator? My specific problem states that there are 8 people inside the elevator and there are 5 floors. In how many ways can they get out if there's a requirement that at least one person leaves at each floor?
The answer is 126000 and I can't get anywhere near it. I do not know if this is the correct answer.
I've tried solving the $x_1 + x_2 + x_3 + x_4 + x_5 = 3$ and then playing with the numbers but didn't get anywhere close.
I am not sure if the order by which people exit the elevator matters as nothing in the problem states that.
 A: The numbers are small, so we can do it by cases. Maybe (Case 1) $4$ people get off at one floor, with the rest $1$ each.  Or maybe (Case 2) it is $3$ on one floor, $2$ another, and the rest $1$ each. Or maybe (Case 3) it is $2$. $2$, $2$ with the rest $1$ each.
I will do Case 1, and Case 3 because it is slightly tricky. You can do Case 2.
Case 1: The floor at which the $4$ people get off can be chosen in $\binom{5}{1}$ ways. For each such choice, the actual people who get off there can be chosen in $\binom{8}{4}$ ways. Once this is done, there are $4$ floors and $4$ people. That gives $4!$ arrangements. So there are
$$\binom{5}{1}\binom{8}{4}4!$$
Case 1 possibilities.
Case 3: We do this one because it is tricky, one can get the wrong answer. The special floors at which the pairs get off can be chosen in $\binom{5}{3}$ ways. Once these have been chosen, the $2$ people who get off at the lowest special floor can be chosen in $\binom{8}{2}$ ways and then the $2$ people who get off at the next special floor can be chosen in $\binom{6}{2}$ ways, and then the $2$ people who get off at the highest chosen floor can be selected in $\binom{4}{2}$ ways. Finally, the two singles can be assigned to the remaining floors in $2!$ ways. Multiply.  
A: Looking at the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 3$ only takes into account the number of people who exit onto each floor, not who the people are. What you want is to partition the set of 8 people into 5 nonempty labelled subsets. That is, you want the number of surjective functions from $\{1,\dots,8\}$ to $\{1,\dots,5\}$. See Stirling numbers of the second kind.
