Difference between revisions of "2000 AIME II Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | Given eight distinguishable rings, let <math>n</math> 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 <math>n</math>. | ||
== Solution == | == Solution == | ||
| + | There are <math>\binom{8}{5}</math> ways to choose the rings, and there are <math>5!</math> distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. 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 <math>\binom {8}{3}</math>. | ||
| + | |||
| + | Multiplying gives the answer: <math>\binom{8}{5}\binom{8}{3}5! = 376320</math>, and the three leftmost digits are <math>\boxed{376}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2000|n=II|num-b=4|num-a=6}} | |
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:13, 12 September 2020
Problem
Given eight distinguishable rings, let 
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 .
Solution
There are 
ways to choose the rings, and there are 
distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. 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 
.
Multiplying gives the answer: 
, and the three leftmost digits are 
.
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
