4. Find the total number of sets of positive integers $(x, y, z)$, where $x, y$ and $z$ are positive integers, with $x<y<z$ such that
$$
x+y+z=203 \text {. }
$$
4. Answer: 3333
Solution. First note that there are $\binom{202}{2}=\frac{202(201)}{2}=20301$ positive integer sets $(x, y, z)$ which satisfy the given equation. These solution sets include those where two of the three values are equal. If $x=y$, then $2 x+z=203$. By enumerating, $z=1,3,5, \cdots, 201$. There are thus 101 solutions of the form $(x, x, z)$. Similarly, there are 101 solutions of the form $(x, y, x)$ and $(x, y, y)$. Since $x<y<z$, the required answer is
$$
\frac{1}{3!}\left(\binom{202}{2}-3(101)\right)=\frac{20301-303}{6}=3333 .
$$