2000 AIME II Problems/Problem 5

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

There are $\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct color arrangements. The number of ways to distribute the rings among the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just $\binom {8}{3}$ .

Multiplying gives the answer: $\binom{8}{5}\binom{8}{3}5! = 376320$ , and the three leftmost digits are $\boxed{376}$ .

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2000 AIME II Problems/Problem 5

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