Let $$A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$$ and $$B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$$. Then the number of elements in the set $$\{(n, m) : n, m \in \{1, 2, \ldots, 10\}$$ and $$nA^n + mB^m = I\}$$ is ______.

We have $$A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$$ and $$B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$$.

$$A^2 = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 4-2 & -4+2 \\ 2-1 & -2+1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} = A$$. Since $$A^2 = A$$, we have $$A^n = A$$ for all $$n \geq 1$$ (that is, $$A$$ is idempotent).

$$B^2 = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1-2 & -2+4 \\ 1-2 & -2+4 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = B$$. Since $$B^2 = B$$, we have $$B^m = B$$ for all $$m \geq 1$$.

Since $$A^n = A$$ and $$B^m = B$$, the equation $$nA^n + mB^m = I$$ becomes $$nA + mB = I$$, namely

$$n\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} + m\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ which gives the system:
$$2n - m = 1 \quad \cdots (1)$$
$$-2n + 2m = 0 \quad \cdots (2)$$
$$n - m = 0 \quad \cdots (3)$$
$$-n + 2m = 1 \quad \cdots (4)$$

From equation (3) we have $$n = m$$. Substituting into (1) gives $$2n - n = 1 \implies n = 1$$, and checking in (4) yields $$-1 + 2(1) = 1$$ ✓.

Therefore, the only solution is $$(n, m) = (1, 1)$$, which lies in $$\{1,2,\ldots,10\}^2$$. Hence, the number of elements in the set is $$\boxed{1}$$.