How many of the numbers, $100,101,\cdots,999$ have three different digits in increasing order or in decreasing order?
$\text{(A) } 120\quad \text{(B) } 168\quad \text{(C) } 204\quad \text{(D) } 216\quad \text{(E) } 240$
For decreasing order, we just need to choose any three digits from the ten: $\tbinom{10}3$
For increasing order, the number cannot start with $0$, so choose from nine: $\tbinom93$
The sum is $204$, giving $\fbox{C}$