Answer: B

Solution:

Let the digits be $a \lt b \lt c$ with $c$ even. Since $a \geq 1,$ no digit is zero, and $c \neq 2$ (there is no room for two smaller nonzero digits).

Once the units digit $c$ is fixed, any two distinct digits below it can be arranged in increasing order in exactly one way. So the count for each $c$ is $\binom{c-1}{2}.$

For $c = 4, 6, 8$ this gives $$\begin{aligned} &\binom{3}{2} + \binom{5}{2} \\ &\quad {}+ \binom{7}{2} = 3 + 10 + 21 \\ &\quad = 34. \end{aligned}$$

Thus, the correct answer is B.