3. Find the number of positive integer solutions to the equation $x_{1}+x_{2}+x_{3}+x_{4}=23$, with the conditions $x_{1} \leqslant 9, x_{2} \leqslant 8, x_{3} \leqslant 7$, $x_{4} \leqslant 6$.
3. Let the equation be $x_{1}+x_{2}+x_{3}+x_{4}=r$, satisfying the conditions $x_{1} \leqslant 9, x_{2} \leqslant 8, x_{3} \leqslant 7, x_{4} \leqslant 6$. The number of positive integer solutions is denoted as $a_{r}$. The generating function for $\left\{a_{r}\right\}$ is $\left(x+x^{2}+\cdots+x^{9}\right)\left(x+x^{2}+\cdots+x^{8}\right)\left(x+x^{2}+\cdots+x^{7}\right)\left(x+x^{2}+\cdots+x^{6}\right)$ $=x^{4} \cdot f(x)=x^{4} \cdot \frac{(1-x)^{9}\left(1-x^{8}\right)\left(1-x^{7}\right)\left(1-x^{6}\right)}{(1-x)^{4}}$. Therefore, finding the coefficient of $x^{2}$ in the expansion, $a_{23}$, is equivalent to finding the coefficient of $x^{19}$ in $f(x)$. Given that $\frac{1}{(1-x)^{4}}=\sum_{r=0}^{\infty} C_{r+3}^{3} x^{r}$, multiplying out the numerator of $f(x)$ and discarding terms with powers higher than 19, the problem reduces to finding the coefficient of $x^{19}$ in $\left(1-x^{6}-x^{7}-x^{8}-x^{9}+x^{13}+x^{14}+2 x^{15}+x^{16}+x^{17}\right) \sum_{r=0}^{\infty} C_{r+3}^{3} x^{7}$. It is easy to see that this coefficient is $C_{22}^{3}-C_{16}^{3}-C_{15}^{3}-C_{14}^{3}-C_{13}^{3}+C_{9}^{3}+C_{8}^{3}+2 C_{7}^{3}+C_{6}^{3}+C_{5}^{3}=115$, which is the desired result.