10. Find the number of positive integer solutions $(x, y, z, w)$ to the indeterminate equation
$$
x+y+z+w=25
$$

satisfying $x<y$.
10. It is known that the number of positive integer solutions to equation (1) is
$$
\mathrm{C}_{24}^{3}=2024 \text {. }
$$

Then the number of positive integer solutions to equation (1) satisfying $x=y$ is the number of positive integer solutions to
$$
2 x+z+w=25
$$

For equation (2), since $z+w \geqslant 2$, we have
$$
1 \leqslant x \leqslant 11 \text {. }
$$

When $x=i(1 \leqslant i \leqslant 11)$, the number of positive integer solutions to $z+w=25-2 i$ is $24-2 i$. Therefore, the number of positive integer solutions to equation (2) is
$$
\sum_{i=1}^{11}(24-2 i)=11 \times 24-2 \sum_{i=1}^{11} i=132 .
$$

Clearly, the number of positive integer solutions to equation (1) satisfying $xy$ is the same. Therefore, the number of positive integer solutions satisfying the problem's conditions is
$$
\frac{2024-132}{2}=946 .
$$