10 Let $x_{1}+x_{2}+x_{3}=x, x_{4}+x_{5}=y, x_{6}=z$, then $x \geqslant 3, y \geqslant 2$, $z \geqslant 1$.

First, consider the indeterminate equation $x+3 y+5 z=21$ satisfying $x \geqslant 3, y \geqslant 2, z \geqslant 1$. Since $x \geqslant 3, y \geqslant 2, z \geqslant 1$, we have
$$
5 z=21-x-3 y \leqslant 12,
$$

then $1 \leqslant z \leqslant 2$.
When $z=1$, we have $x+3 y=16$, the positive integer solutions of this equation satisfying $x \geqslant 3, y \geqslant 2$ are
$$
(x, y)=(10,2),(7,3),(4,4) .
$$

When $z=2$, we have $x+3 y=11$, the positive integer solutions of this equation satisfying $x \geqslant 3, y \geqslant 2$ are $(x, y)=(5,2)$.

Therefore, the positive integer solutions of the indeterminate equation $x+3 y+5 z=21$ satisfying $x \geqslant 3, y \geqslant 2, z \geqslant 1$ are
$$
(x, y, z)=(10,2,1),(7,3,1),(4,4,1),(5,2,2) .
$$

Since the number of positive integer solutions of the equation $x_{1}+x_{2}+x_{3}=x(x \in \mathbf{N}, x \geqslant 3)$ is $\mathrm{C}_{x-1}^{2}$, and the number of positive integer solutions of the equation $x_{4}+x_{5}=y(y \in \mathbf{N}, x \geqslant 2)$ is $\mathrm{C}_{y-1}^{1}$, by the principle of multiplication, the number of positive integer solutions of the original indeterminate equation is
$$
\mathrm{C}_{9}^{2} \mathrm{C}_{1}^{1}+\mathrm{C}_{6}^{2} \mathrm{C}_{2}^{1}+\mathrm{C}_{3}^{2} \mathrm{C}_{3}^{1}+\mathrm{C}_{4}^{2} \mathrm{C}_{1}^{1}=36+30+9+6=81
$$
10 (15 points) Find the number of positive integer solutions to the indeterminate equation $x_{1}+x_{2}+x_{3}+3 x_{4}+3 x_{5}+5 x_{6}=21$.