Using exclusion and inclusion method to calculate number of functions

Question

Let $A =\{1,2,3,4,5,6\}$.

I want to find the number of functions $f : A \rightarrow A$ such that

$|f^{-1}(i)|=i$ for every $i\in \{1,2,3\}$.

anyone has an idea, I tried a lot but didn't get to the answer.

Thank you

Answer 1

You want $f$ to map one value into $1$, two values into $2$ and three values into $3$.

$f^{-1}[\{1\}]$ is a singleton containing any value between $1$ and $6$.

$f^{-1}[\{2\}]$ is a set containing any two values taken from $\{1,\ldots,6\} \setminus f^{-1}[\{1\}]$.

$f^{-1}[\{3\}]$ is the uniquely determined set given by $\{1,\ldots,6\} \setminus (f^{-1}[\{1\}] \cup f^{-1}[\{2\}])$.

Hence, the total number of possibilities is $\binom{6}{1} \cdot \binom{5}{2} \cdot \binom{3}{3} = 60$.