The bigger digits

We can choose the 3 distinct digits out of the 9 digits allowed (1, 2, ..., 8, 9):

${ 9 \choose 3 } = 84 \text { ways. }$

Since we have to put the largest digit in the middle and we can arrange the remaining 2 digits in 2! = 2 ways, therefore our answer should be:

$84 × 2 = \boxed {168}$