The number of 3-digit numbers we of the form $xyz$ with $x < y, z < y$ and $x \neq 0$ is

Let the 3-digit number be $xyz$, where $x, y, z \in \{0, 1, 2, \dots, 9\}$.

Since $x \neq 0$, $x \in \{1, 2, \dots, 9\}$.

Given $x < y$ and $z < y$, the minimum value of $y$ must be $2$ because $x \ge 1$. Thus, $y \in \{2, 3, \dots, 9\}$.

For a fixed value of $y$, $x$ can take values from $1$ to $y-1$, which gives $y-1$ choices. The digit $z$ can take values from $0$ to $y-1$, which gives $y$ choices.

The number of possible 3-digit numbers for a given $y$ is $(y-1)y$.

The total number of such 3-digit numbers is the sum over all possible values of $y$:

$\sum_{y=2}^{9} y(y-1) = \sum_{y=1}^{9} (y^2 - y)$

$\sum_{y=1}^{9} y^2 - \sum_{y=1}^{9} y = \dfrac{9 \times 10 \times 19}{6} - \dfrac{9 \times 10}{2}$

$285 - 45 = 240$

Answer: $240$