Theorem: For any $n,r \in \mathbb{N}$ such that $0 \le r \le n$, $\sum_{k=r}^{n} \binom{k}{r} = \binom{n+1}{r+1}$ Proof: I will proceed with a proof by double counting. Consider the set $K := \{S \subseteq \{0,1,2,3, \ldots, n\}: |S| = r\}.$ The number of subsets containing $r+1$ elements is $\binom{n+1}{r+1}$. Thus, $|K| = \binom{n+1}{r+1}$. Now, consider counting $|K|$ by selecting a maximum value of $S$. In order to have $|S| = r$, we must have that $M \ge r$ otherwise there are only $r$ choices to create an $r+1$ subset of $\{0,1,2,\ldots, n-1\}$ which is not possible. For each $r'$ such that $r \le r' \le n$, we will have $\binom{r'}{r}$ choices of elements to put in $S$. Therefore, $ |K| = \binom{r}{r}+\binom{r+1}{r}+\ldots+\binom{n-1}{r}+\binom{n}{r} = \sum_{k = r}^{n} \binom{k}{r} .$ In sum, we find that $\sum_{k=r}^{n} \binom{k}{r} = \binom{n+1}{r+1}$ and so this means that $ SUM = r!\sum_{k=r}^{n} \binom{k}{r} = r!\binom{n+1}{r+1} .$