1. The problem asks for the coefficient of $x^3$ in the expansion of $(1 + x)^5$. 2. We use the binomial theorem, which states: $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ 3. Here, $a = 1$, $b = x$, and $n = 5$. 4. The term containing $x^3$ corresponds to $k = 3$: $$ \binom{5}{3} 1^{5-3} x^3 = \binom{5}{3} x^3 $$ 5. Calculate the binomial coefficient: $$ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{20}{2} = 10 $$ 6. Therefore, the coefficient of $x^3$ is $10$. Final answer: 10