Dare to do 2

The number of positive integer solutions of $x+y+z=6$ is $(_{ 3-1 }^{ 6-1 }{ ) }=10$ .

As $x+y+z=6, u+v\le 9$

$u+v=k$ can only have positive integer solutions when $k\ge 2$ . Thus there are 8 possible values of $k$ , namely $2\le k \le 9$ .

Observe that the number of positive integer solutions to $u+v=k$ where $2\le k\le 9$ is $\sum _{ i=2 }^{ 9 }{ { ( }_{ 2-1 }^{ i-1 } } )=36$ .

There are 10 possible values of $x+y+z$ per each value of $u$ and $v$ , hence the number of possible positive integer solutions is $10\cdot 36=360$