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, 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>. | + | 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>. | 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 16:03, 24 June 2011
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 |