Q:

Number of ways of choosing $x,y,z$ such that $z>max(x,y)$

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?

A:

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 $2\cdot C(10,3)+(10,2)$.

