In a certain question I am asked to find how many possible subsets $A,B \subset {\{1 \dots n\}}$ are there, given that the intersection of $A$ and $B$, $A \cap B$ must have one element. There is no neccessity that $A \cup B = \{1 \dots n\}$ and it is supposed that $A,B$ are distinguishable. Here is my reasoning:
To begin with, one has to see what element is in the intersection: there are $n$ options. Therefore, there are $n-1$ remaining numbers. For each number in that new set, there are three options:
- The number $i$ is in $A$
- The number $i$ is in $B$
- The number $i$ is neither in $A$ nor $B$
Therefore, there are $3^{n-1}$ options, and using the product rule, the final result is of $n \cdot 3^{n-1}$ possibilities. Does this look sound?