Question: Let $A$ be a group of sequences of length 9 made of $\{0,1\}$ and its given that $|A|=52$. Show that there exists 2 sequences $a_1$, $a_2$ that belong to $A$ so you could replace at most 2 elements in $a_1$ and get $a_2$.
Tried a few ways to define pigeons and pigeonholes but it I can't get it to what you actually need to prove. tried taking pigeons as all possibles couples ${52} \choose {2}$ possibilities but get stuck on the pigeonholes. I'm thinking maybe a double pigeonhole principle is needed but can't get to how to define it.
Hint: For each string $a\in A$, let $S_a$ consist of the 9 strings formed by changing a single letter in $a$, together with $a$ itself. That is, $|S_a|=10$. If $a=000000000$, then $S_a$ would contain $a$, together with the $9$ strings that have a single $1$.
If $S_a$ and $S_b$ have string in common, then $a$ and $b$ would differ in two spots. So if no two strings differ in two spots, then the sets $S_a$ are all disjoint...