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Let $X$ be a set with $n$ elements. Prove that $$ \sum_{Y, Z \subseteq X}|Y \cap Z|=n \cdot 4^{n-1} $$ The sum is over all possible pairs $(Y, Z)$ of subsets of $X$.

I do not need a solution, but rather I need help in understanding this problem, for trying to understand what it says, I took $n=3$ case, say $X=\{1,3,5\}$, now we can list out all possible subsets that are as following (excluding null set as its not relevant here):

$$A\{1\}, B\{3\}, C\{5\}, D\{1,3\}, E\{1,5\}, F\{3,5\}, G\{1,3,5\}$$

Next I list out all possible non empty intersections of these subsets, which is as following:

$$\begin{aligned} & A \cap D=\{1\}, \quad A \cap E=\{1\}, \quad A \cap G=\{1\} \\ & B \cap D=\{3\}, \quad B \cap F=\{3\}, \quad B \cap G=\{3\} \\ & C \cap E=\{5\}, \quad C \cap F=\{5\}, \quad C \cap G=\{5\} \\ & D \cap E=\{1\}, \quad D \cap F=\{3\}, \quad D \cap G=\{1,3\} \\ & E \cap F=\{5\}, \quad E \cap G=\{1,5\}, \quad F \cap G=\{3,5\} . \end{aligned}\\$$

Clearly the sum of cardinality of all these subsets is $18$, and not $3 \times 4^{2}=48$, so what an I misunderstanding here?

Asaf Karagila
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OlympiadRunner
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    It looks to me like you're leaving out two things: (1) the pairs that you've listed but in the opposite order $(D\cap A,$ for example, in addition to $A\cap D$); and (2) pairs with $Y=Z,$ such as $A\cap A.$ – Mitchell Spector Aug 26 '24 at 09:57
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    Since $(Y,Z)$ is ordered and allows repetitions, your list is missing some possibilities, such as $A\cap A={1}$ and $D\cap A={1}$. – Sangchul Lee Aug 26 '24 at 09:58
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    Welcome to MathStackEx , +1 for nicely worded question showing your effort , while high-lighting your confusion here ! – Prem Aug 26 '24 at 11:24

1 Answers1

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We have to "reverse-engineer" the matter here.

We have $48 = 12 + 18 \times 2$ : I am thinking out loud :

  • We can see that $12$ matches the list given with $7$ Subsets. Why is that number coming here ?
  • Maybe $18$ had to be counted twice ? Why ?

Solution to thinking out loud:

  • We have no restrictions on "uniqueness" : $Y = Z$ , $Y \not = Z$ : We have to consider both.
  • We have no restrictions on "order" : $(Y,Z)$ , $(Z,Y)$ : We have to count twice due to renaming.

That will make the question match the answer.

Prem
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