PK 2015 Q22

Since the three digits \(a\), \(b\), \(c\) are different and \(\overline{ab} < \overline{bc} < \overline{ca}\), it is necessary that \(a < b < c\). But this condition is also sufficient. Also note that none of \(a\), \(b\), \(c\) are zero because they each represent the tens digit of a two-digit number.
There are 9 ways to pick a non-zero digit, 8 ways to pick a second (different) digit, and 7 ways to pick a third digit. These digits can be arranged in \(3 \times 2 \times 1 = 6\) ways, but only one of these will be in ascending order (we need \(a < b < c\)). Hence there are \((9 \times 8 \times 7) \div 6 = 84\) ways to choose the digits \(a\), \(b\), \(c\).