Question

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 $n.$

Answer

Solution:

Choose which five rings to use in $\binom{8}{5} = 56$ ways, and order them (reading down the first finger, then the second, and so on) in $5! = 120$ ways. It remains to split the ordered list into four possibly empty consecutive blocks, one per finger: the number of compositions of $5$ into $4$ nonnegative parts, which by stars and bars is $\binom{8}{3} = 56.$

Therefore $n = 56 \cdot 120 \cdot 56 = 376320,$ whose leftmost three nonzero digits are $376.$

2000 AIME II Problem 5

Solution:

Problem 5 in Other Years