Difference between revisions of "2008 AMC 12A Problems/Problem 21"
(correct problem) |
|||
Line 17: | Line 17: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}} | {{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}} | ||
− | |||
− |
Revision as of 22:38, 19 February 2008
Problem
A permutation of is heavy-tailed if . What is the number of heavy-tailed permutations?
Solution
There are total permutations.
For every permutation such that , there is exactly one permutation such that . Thus it suffices to count the permutations such that .
, , and are the only combinations of numbers that can satisfy .
There are combinations of numbers, possibilities of which side of the equation is and which side is , and possibilities for rearranging and . Thus, there are permutations such that .
Thus, the number of heavy-tailed permutations is .
See also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |