How many have there pairs of disjoint subsets of a set of $n$ elements?

Question

-How many have there pairs of disjoint subsets of a set of $n$ elements?

I did other assignments about pairs of subsets where one is a subset of the second, using functions similar to characteristic functions. But I dont have an idea about this one.

Answer 1

Assuming you mean ordered pairs:

Step 1: Choose a subset.

Step 2: Choose a subset from the remaining (unchosen) elements.

If you restrict your attention to the situation in which the first subset has size $m$, there are $\binom{n}{m}$ ways to choose the first subset, and $2^{n-m}$ ways to choose the second given the first. (This assumes that the empty set is allowable.) So, the number you want would be $$ \sum_{m=0}^{n}\binom{n}{m}2^{n-m}. $$ How can we simplify this? Well, we can write $$ \sum_{m=0}^{n}\binom{n}{m}2^{n-m}=2^n\sum_{m=0}^{n}\binom{n}{m}\left(\frac{1}{2}\right)^m=2^n\left(1+\frac{1}{2}\right)^n=3^n. $$

That might seem surprising; but there's another way we could've gotten it. To every one of your $n$ elements, assign a label $0$, $1$, or $2$. The ones with $0$ are left out; the ones with $1$ occur in the first subset; and the ones labeled $2$ occur in the second subset. This is a bijection which shows that there are as many arrangements as there are functions from $n$ objects to $\{0,1,2\}$, which is also precisely $3^n$.

Answer 2

For each subset $A$ with $k$ elements in the set $X$ with $n$ elements, a paired subset is any subset $B\subset \mathcal P(X-A)$, which has $2^{n-k}$ elements. Hence the number of pairs of disjoint subsets of $X$, taking the order into account, is: $$\sum_{k=0}^n\binom nk 2^{n-k}=\sum_{k=0}^n\binom nk 2^k=3^n.$$ If you do not take the order into account, as each pair is obtained twice, this number becomes $\;\dfrac{3^n}2$.