Here is a combinatorial problem: let $\Sigma=\{\alpha_1,\ldots,\alpha_n\}$ be an alphabet and we consider any words over $\Sigma$ of length $n$. We also define over the set of such words the Hamming distance: $$d(\omega,\omega')=\#\{i\in \{1,\ldots,n\} \vert a_i\neq b_i\}$$ where $\omega=a_1\ldots a_{n}$ and $\omega'=b_1\ldots b_n$ are two words.
The questions are the following:
Let $\omega_0$ be the word built with only one letter: $$\omega_0=\underbrace{\alpha\ldots \alpha}_{n \text{ times}}$$ what is the cardinality of a biggest set $\Omega(\omega_0)$ of words of length $n$ containing $\omega_0$ and such that $\forall \omega,\omega'\in \Omega(\omega_0), \ d(\omega,\omega')=n-1$ ?
What if we ask the same question for words of length $n+1$ where $\omega_0=\underbrace{\alpha\ldots \alpha}_{n+1 \text{ times}}$ and $\Omega(\omega_0)$ is a maximal set of words of length $n+1$ containing $\omega_0$ such that $\forall \omega,\omega'\in \Omega(\omega_0), \ d(\omega,\omega')=n-1$?
It seems that the result is the same and depends only on the cardinality of the alphabet but I don't know how to prove it neatly. Any ideas?