# Inclusion exclusion and partition of a set - making sure I understand the concepts

If I may, I would like to verify my solution of a couple of homework questions, and by doing so asking a few questions about these topics.

1. Let $$X$$ be a set of size $$n$$. How many distinct triplets $$(A, B, C)$$ are there such that A, B and C are non-empty disjoint subsets of X? (I hope I'm using the right terminology)

My idea was to associate a function from $$X$$ to $$\{0, 1, 2, 3\}$$ so that for each $$x \in X$$, for example if $$x \in A$$ then $$f(x) = 1$$, etc... and if $$x$$ is not element of neither $$A$$, $$B$$ nor $$C$$ then $$f(x) = 0$$. So in total without restrictions there are $$4^n$$ such functions, we'll call that group $$U$$.

Since $$A, B, C$$ must not be empty, I used inclusion-exclusion to reach this answer: $$4^n - 3 \cdot 3^n + 3 \cdot 2^n - 1$$ Because there are $$3^n$$ functions where one of $$A, B, C$$ is empty, $$2^n$$ where two of them are empty and one function where they are all empty. Is this right?

1. How many ways are there to select $$3$$ non-empty disjoint subsets of $$X$$ if they are non-distinguishable?

Here I'm less certain of my answer and also it isn't simplified enough. Also, Stirling numbers are not included in my course subjects.

We can choose to partition all the $$n$$ elements to $$3$$ sets, there are $$S(n, 3)$$ to do that. We can choose $$n-1$$ elements out of $$n$$ and partition them to $$3$$ sets, so $$\frac{n(n-1)(n-2) \cdot S(n-1, 3)}{6}$$ ways, and so on. The answer is the sum of this series.

Is this solution correct? How to simplify this series sum? What is a simpler solution, maybe without using Stirling number?

Thank you.