Problem: Find the number of ordered quadruples $\left(x_1, x_2, x_3, x_4\right)$ of odd positive integers such that $x_1+x_2+x_3+x_4=98$. Solution: Write $x_{i}= 2k_{i} + 1$ for each $i \in \{ 1,2,3,4 \}$ where $k_{i} \in \mathbb{Z}_{\geq_{0}}$. Then, $x_{1}+x_{2}+x_{3}+x_{4} = 2(k_{1}+k_{2}+k_{3}+k_{4}) + 4 = 98$. We find that $k_{1}+k_{2}+k_{3}+k_{4} = 47$. The number of solutions to this is a classic distributions problem and the answer is $\binom{47 + 4 - 1}{4 - 1} = \binom{50}{3} = 50\cdot 49 \cdot 48 \cdot \frac{1}{3!} = 19600$.