1. The problem asks for the coefficient of $x^2 y^3$ in the expansion of $(x + y)^5$. 2. Use the binomial theorem: $$(x + y)^5 = \sum_{k=0}^5 \binom{5}{k} x^k y^{5-k}$$ 3. We want the term where the power of $x$ is 2 and the power of $y$ is 3, so $k=2$. 4. The coefficient is given by the binomial coefficient $\binom{5}{2}$. 5. Calculate $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$. 6. Therefore, the coefficient of $x^2 y^3$ in the expansion is **10**. Final answer: 10