Question:

We need 3-digit numbers $xyz$ where $x < y$, $z < y$, and $x \neq 0$. For a fixed middle digit $y$, we count valid choices for $x$ and $z$ independently. For each value of $y$: $x$ can be chosen from $\{1, 2, \ldots, y-1\}$ (giving $y-1$ choices, excluding 0), and $z$ can be chosen from $\{0, 1, \ldots, y-1\}$ (giving $y$ choices). Thus for each $y$, there are $(y-1) \cdot y$ valid numbers. Summing over $y = 1$ to $9$: $\lambda = \sum_{y=1}^{9} y(y-1) = \sum_{y=1}^{9} (y^2 - y) = \frac{9 \cdot 10 \cdot 19}{6} - \frac{9 \cdot 10}{2} = 285 - 45 = 240$.