2008 AMC 12A Problems/Problem 21
Problem
A permutation of is heavy-tailed if . What is the number of heavy-tailed permutations?
Solution 1
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 .
Solution 2 (Casework)
We use case work on the value of .
Case 1: . Since , can only be a permutation of or . The values of and , as well as the values of and , are interchangeable, so this case produces a total of solutions.
Case 2: . Similarly, we have is a permutation of , , or , which gives a total of solutions.
Case 3: . is a permutation of or , which gives a total of solutions.
Case 4: . is a permutation of , , or , which gives a total of solutions.
Case 5: . is a permutation of or , which gives a total of solutions.
Therefore, our answer is .
-MP8148
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.