1. The problem is to evaluate the binomial coefficient $\binom{2}{3}$. 2. The binomial coefficient $\binom{n}{k}$ is defined as $\frac{n!}{k!(n-k)!}$ where $n!$ is the factorial of $n$. 3. Important rule: If $k > n$, then $\binom{n}{k} = 0$ because you cannot choose more elements than available. 4. Here, $n=2$ and $k=3$, and since $3 > 2$, we have $\binom{2}{3} = 0$. 5. Therefore, the final answer is: $$\binom{2}{3} = 0$$