Let $x,y,z$ belong $\{1,2,3,....,10\}$ then find the number of ways of choosing $x,y,z$ such that $z>max(x,y)$ .
The selection of $z$ always depends on the selection on $x,y$ so making cases is the only way or is there any smart approach to solve this problem?
You can first choose the numbers and then label them. For example choose three numbers, say $2,5,7$. You can set $z=7$ and then you have two choices. Either $x=2$, $y=5$ or $x=5$, $y=2$. Another case is that you choose two numbers $4,6$, set $z=6$ and $x=y=4$. Therefore the answer is $2\cdot C(10,3)+(10,2)$.