5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups.
5. 336005 .

The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}-3 \times 1005}{3!}=336005 \text { (sets). }
$$