HMMT February

Let $f(\{1,2,3\})$ be $A$. Then $A \cap f(A)=\emptyset$, so $A$ must be a subset of $\{4,5\}$. If $B=\{4,5\}$, there are $2^{3}-2$ ways to assign each element in $\{1,2,3\}$ to a value in $\{4,5\}$, and 9 ways to assign each element of $\{4,5\}$ to a value in $\{1,2,3\}$, for a total of 54 choices of $f$. If $A=\{4\}$, there is 1 possible value for each element of $\{1,2,3\}, 4$ ways to assign $\{4\}$ with a value from $\{1,2,3,5\}$, and 5 ways to assign a value to $\{5\}$. Similarly, if $A=\{5\}$, there are $4 \cdot 5=20$ choices for $f$. In total, there are $54+20 \cdot 2=94$ possible functions.