A general formula for $\sum (k-1)(k-2)(k-3)$? What is a "simpler" formula for
$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$

Perhaps a messy and boring way, we can use the generating function. 
$$\sum_{k=3}^{n}\frac{(k-1)(k-2)(k-3)}{6}=\frac{1}{6}\sum_{k=2}^{n-1}k(k-1)(k-2)$$
In addition, the generating function for $k(k-1)(k-2)$ is $x^3\left(\frac{1}{1-x}\right)^{(3)}.$
Hence, the sum is the coefficient of $x^{n-1}$ in $\frac{1}{6}\frac{1}{1-x}x^3\left(\frac{1}{1-x}\right)^{(3)}=\frac{x^3}{(1-x)^5}$, which is 
$$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$
