Two partitions of $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ I recently stumbled upon the following problem, and I have no idea how to proceed. 
Let $S=\{1, 2, 3, 4, 5, 6, 7, 8, 9 \}.$ Let $P_1, P_2$ be partitions of $S$. For $x \in S$, let $\pi_1(x)$ be the number of elements of the set $A \in P_1$ for which $x \in A$, and let $\pi_2(x)$ be the number of elements of the set $B \in P_2$ for which $x \in B$. Show that there exist distinct $x, y \in S$ with $\pi_1(x) = \pi_1(y)$ and $\pi_2(x) = \pi_2(y)$. 
I know that some pigeon-hole principle type of argument is involved, I tried to visualize the problem in lots of ways, but I can't seem to find a solution without the number of things to be checked getting extremely large. A hint would be appreciated. 
 A: There can be at most three different part sizes in a partition of $[9]$. (The smallest partition with $4$ different part sizes would be $1+2+3+4$, with $10$ elements.) Also, it's impossible to have three part sizes with exactly $3$ elements for each part size (as that's only possible for part sizes $1$ and $3$). Thus, there is at least one part size in $P_1$  with at least $4$ elements in parts of that size. By the pigeonhole principle, at least two of these $4$ elements must be in parts with the same size, since there are at most $3$ part sizes in $P_2$.
A: Let $\pi(x) = (\pi_1(x),\pi_2(x))$. We want to show that $\pi$ is not injective.
So assume $\pi$ is injective. Consider the possible distinct values of $\pi_1$, let's call them $V_1 = \pi_1(S)$ and $V_2 = \pi_2(S)$.
$|V_1|$ can be at most $3$, since if there were $4$ or more sizes of partition classes, there would need to be at least $1+2+3+4=10$ elements in $S$. By symmetry also $|V_2| \leq 3$.
Since $\pi_1$ and $\pi_2$ each have at most three distinct values, $\pi$ can have at most $9$ distinct values. For $\pi: S \to V_1 \times V_2$ to be injective, we must have $|V_1| = |V_2| = 3$ and $\pi$ must be bijective. This implies that $|\pi_1^{-1}(v_1)| = |V_2| = 3$ for all $v_1 \in V_1$.
Now since $\pi_1^{-1}(v_1)$ is the disjoint union of sets of size $v_1$, we have that $v_1$ divides $|\pi_1^{-1}(v_1)|$, so $v_1$ can only be $1$ or $3$. Hence we have $V_1 \subseteq \{1, 3\}$ and therefore $3 = |V_1| \leq |\{1, 3\}| = 2$. By contradiction, $\pi$ is not injective.
