Application of combination with repetition

When x=1, there are 5 solutions for positive y, z such that y+z=6, When x=2, there are 4 solutions for positive y, z such that y+z=5, and so on to x=5 So the answer is 5+4+3+2+1=15

The simplest approach to solve this problem is to use stars & bars technique. The desired sum $7$ can be represented using $7$ stars. Each combination of $x, y$ and $z$ that adds up to $7$ can be represented by putting two 'bar' symbols in empty places between the stars. There are $6$ places between the $7$ stars and you need to choose $2$ of them where the bars will be put. There is ${6 \choose 2} = 15$ ways of doing this and thus $15$ distinct integer triples that add up to $7$ .

no. of positive integer solution $= _6C_2=\frac{6!}{4!2!}=15$