2000 AIME II Problems/Problem 5
Contents
[hide]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
.
Solution 2
There are to choose the rings.
Now, call the rings
,
,
,
, and
. Any possible ring combination can be generating by scrambling
and adding three dividers - the things to the left of the first divider will be on the first finger, the things between the first and the second will be on the second finger, and so on. For example,
represents ring
on the first finger,
on the second, none on the third, and
,
, and
in that order on the fourth. In other words, we simply need to count the number of distinguishable rearrangements of
. There are
of them.
Thus the total number of ways is and we submit
.
~ NamelyOrange
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 | ||
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