Difference between revisions of "2000 AIME II Problems/Problem 5"
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== Solution == | == Solution == | ||
− | There are <math>\binom{8}{5}</math> ways to choose the rings, <math> | + | There are <math>\binom{8}{5}</math> ways to choose the rings, and there are <math>5!</math> 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 <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>. | 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 == | ||
{{AIME box|year=2000|n=II|num-b=4|num-a=6}} | {{AIME box|year=2000|n=II|num-b=4|num-a=6}} | ||
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+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 15:57, 30 August 2008
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 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 .
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 |