How can we say that the number of disjoint subsets of a set containing n elements is 1+ $\frac{3^{n} - 1}{2}$

Question

I am aware that each element of a set with cardinality n will have three ways of occupation: either in the parent set, or in either of its disjoint subsets. Hence, we have $3^{n}$ possibilities. But since we counted a possibility of both the disjoint sets being null, and hence equal, we subtract 1. In addition, either of the disjoint subsets are interchangeable, so we counted twice and hence we divide by 2. But don't understand why we add another 1 to this whole expression.

Answer 1

The case of two empty sets should be counted because they are disjoint. When we subtracted the $1$ before the division by $2$ we removed that case because we did not count it twice. We add it back in at the end.

Answer 2

We can count the ordered pairs $(A,B)$ of disjoint subsets of $\mathbf{n}=\{1,\ldots, n\}$ by counting all functions $f: \mathbf{n} \to \{a,b,n\}$ where $f(i) = a$ means $i \in a$, $f(i) = b$ means $i \in b$ and $f(i) = n$ that $i \notin (A \cup B)$. There are of course $3^n$ many such functions.

The constant function with value $n$ then corresponds to $(\emptyset,\emptyset)$, which is the only pair that stays the same when interchanged, so we disallow that function ($3^n -1$ are left, nicely even), and for all others we count all disjoint doubletons $\{A,B\}$ (which is what we really want to count) unequal to a singleton $\{\emptyset\}$ exactly twice because $\{A,B\}$ can be a "flattened form" of $(A,B)$ or $(B,A)$. So all the number of real doubletons are given by $\frac{3^n -1}{2}$. The $+1$ is for wanting to also count two empty sets as a solution too, which we don't get from a non-"constant $n$" function.