Combinatorics - Sorting objects into specific sets with no empty set using a generating function How can you sort 90 people in 3 different rooms with no room being empty?
With no restrictions sorting 90 people in 3 rooms would be 3^90. When it comes to no room empty, I start with (x+x^2/2!+...)^3 and get lost.
The simplest approach is to use an inclusion-exclusion argument. There are, as you say, $3^{90}$ unrestricted distributions; from that you want to subtract the distributions that leave at least one room empty. For each of the $3$ rooms there are $2^{90}$ distributions leaving that room empty, so to a first approximation there are $3\cdot 2^{90}$ distributions that leave at least one room empty. However, this counts every distribution that leaves two rooms empty twice, once for each of the two empty rooms. Plainly there are $3$ such distributions: if two rooms are empty, all $90$ people are in one of the $3$ rooms. Thus, there are really only $3\cdot 2^{90}-3$ distributions that leave at least one room empty, and the final answer is therefore $3^{90}-3\cdot 2^{90}+3$.
If you consider the series
$$(x+x^2+x^3+\cdots)^3$$
coefficient of $x^{90}$ counts the number of ordered triplets $(k_1, k_2, k_3)$ of positive integers such that $k_1+k_2+k_3=90$. This counts the number of ways to assign 90 people into three rooms provided that the rooms are labeled.
If the rooms are not labeled and you only care about the number of people in each room, you are asking for a type of restricted partition which has a more complicated derivation.