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>\binom{8}{3}</math> ways to distribute the rings among the fingers, and <math>5!</math> distinct color arrangements. |
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| + | Multiplying gives the answer: <math>\binom{8}{5}\binom{8}{3}5! = 376320</math>, and the three leftmost digits are <math>\boxed{376}</math>. | ||
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{{AIME box|year=2000|n=II|num-b=4|num-a=6}} | {{AIME box|year=2000|n=II|num-b=4|num-a=6}} | ||
Revision as of 20:33, 18 March 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, 
ways to distribute the rings among the fingers, and 
distinct color arrangements.
Multiplying gives the answer: 
, and the three leftmost digits are 
.
| 2000 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 6 | |
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