This question was inspired by the binomial theorem for rings. For commutative rings, we have the identity
$(a+b)^n = \sum_{k=0}^n {n \choose k}a^kb^{n-k}$
which does not hold for non-commutative rings. However, we can still expand $(a+b)^n$ to get \begin{align*} (a+b)^2 &= a^2 + ab + ba + b^2 \\ (a+b)^3 &= a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 \\ \cdots \\ (a+b)^n &= aa\ldots a + aa \ldots ab + aa \ldots ba + aa \ldots bb + \ldots \end{align*}
This motivates the following question. For every $n$, is it possible to find a ring $R$ with elements $a$ and $b$ such that the $2^n$ terms in the expansion of $(a+b)^n$ are distinct?
There is also a stronger question: is it possible to find $a$ and $b$ such that for every $n$, this is true?
For example, taking $A = \begin{bmatrix}2 &1 \\0 &1\end{bmatrix}$ and $B = \begin{bmatrix}1 &2 \\0 & 1\end{bmatrix}$ from $M_2(\mathbb{Z})$ works for $n = 1, 2, 3$ and $4$ but fails for $n = 5$ because $ABBBA = BBAAB$.