Compute $\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4}$.
I know that $\dbinom{n}{k}$ is the number of ways to choose k objects out of n distinct objects, regardless of order. So $\dbinom{4}{0}$ is the number of ways to choose 0 objects out of 4, which is just 1. Similarly, $\dbinom{4}{4}$ is the number of ways to choose 4 objects out of 4, which is also 1. For the other terms, I can use the formula $\dbinom{n}{k}=\frac{n!}{k!(n-k)!}$, where $n!$ is the product of all positive integers up to n. For example, $\dbinom{4}{1}=\frac{4!}{1!3!}=\frac{4\times 3\times 2\times 1}{1\times 3\times 2\times 1}=\frac{4}{1}=4$. Similarly, $\dbinom{4}{2}=\frac{4!}{2!2!}=\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=\frac{6}{1}=6$ and $\dbinom{4}{3}=\frac{4!}{3!1!}=\frac{4\times 3\times 2\times 1}{3\times 2\times 1\times 1}=\frac{4}{1}=4$. So the sum I need to compute is $1+4+6+4+1=16$. 