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Difference between revisions of "2000 AIME II Problems/Problem 5"

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

Revision as of 18:09, 20 December 2008

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 arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger, 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 $\binom {8}{3}$.

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

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
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