Problem: Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$. Solution: Pick the rings. There are $\binom{8}{5}$ ways to do this. Order the rings. There are $5!$ ways to do this. We are considering strings now. Insert $3$ dividers to seperate this ordering into four groups. All rings to the left of the left-most divider are for finger 1, all rings between the leftmost and penultimate leftmost are for finger two, and so on. Critically, for any ring arrangement, we can always point to a string like this that represents the arrangement, and it is impossible for two different strings arrangements to correspond to the same ring arrangement. Hence, the total number of arrangements is $\binom{8}{5}5! \binom{8}{3} = 376320$ and so the answer is $376$.