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**Soroban** Hello, shaoen01!

Suppose $\displaystyle n \:=\:30 \:=\:2\cdot3\cdot5$

With three factors to choose from, there are: $\displaystyle 2^3 = {\bf8} $ possible subsets.

They are: .$\displaystyle \{\;\},\;\{2\},\;\{3\},\;\{5\},\;\{2,3\},\;\{3,5\} ,\;\{2,5\},\;\{2,3,5\} $

But the question is: how many two-set partitions are there?

There are **four**: .$\displaystyle \begin{array}{cc}\{\;\} &\{2,3,5\} \\ \{2\} & \{3,5\} \\ \{3\} & \{2,5\} \\ \{5\} & \{2,3\} \end{array}\quad \text{ The products are: }\begin{Bmatrix}1\cdot30 \\ 2\cdot15 \\ 3\cdot10 \\ 5\cdot6\end{Bmatrix}$

The partition $\displaystyle \{2\}\;\;\{3,5\}$ is the same as $\displaystyle \{3,5\}\;\;\{2\}$

. . because $\displaystyle 2\cdot15\text{ and }15\cdot2$ are the same factoring.