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How many pairs $(A_1, A_2)$ of subsets of $\{1,2,\ldots,n\}$ are there such that $A_1 \cap A_2 = \emptyset$?

I am to give a solution involving binomial coefficients.

The hint I was given is to choose $A_1$ and then choose $A_2$ as a subset of its complement. Then determine the sum of products of binomial coefficients.

I do not know where to begin, so any guidance is appreciated.

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  • $\begingroup$ If you want the solution specifically in terms of binomials, I'd say $\sum_{k=0}^n 2^k\binom nk$, though I'd rather call it $3^n$. $\endgroup$ – Gae. S. Aug 2 '19 at 19:24
  • $\begingroup$ Well, not sure that hint is optimal but for another approach: any such such selection is the same as a map to $\{1,2,3\}$ where an element maps to $1$ if it is in $A_1$, $2$ if it is in $A_2$, and $3$ if it is in neither. $\endgroup$ – lulu Aug 2 '19 at 19:25
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Take any $k \in \{ 0,...,n \}$ and say that set $A_1$ will have $k$ elements chosen from $\{1,...,n\}$. Then there are $n-k$ elements left, and set $A_2$ can be any subset of set of those remaining elements. That is, for a fixed $k$, we have ${n \choose k}$ ways of choosing $A_1$, then for every $A_1$ (cause this is only matter of $|A_1|$) we have $2^{n-k}$ ways to choose $A_2$. Now we have to sum over all values $k$, which gives us: $\sum_{k=0}^n {n \choose k} 2^{n-k}$.

However, we can count it in different way. Note, that since we want $A_1 \cap A_2 = \emptyset$, that means we can form a set $A_3$ (for every choice of $A_1,A_2$) containing only those elements, which are in set $\{1,...,n\}$ and aren't in $A_1$ nor $A_2$. So we can say, we're just considering triples $(A_1,A_2,A_3)$, where $A_i \cap A_j = \emptyset$, for every $i,j \in \{1,2,3\} $, $ i \neq j$. And $A_1 \cup A_2 \cup A_3 = \{1,...,n\}$. How to do this? Just take elements in $\{1,...,n\}$ one by one, and ask yourself, whether u want to put it in $A_1,A_2$ or $A_3$. How many ways it gives? Well, there are $n$ elements, and $3$ choices for every, so $3^n$.

We ended up with $\sum_{k=0}^n {n \choose k} 2^{n-k} = 3^n$

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