Difference between revisions of "2008 AMC 12A Problems/Problem 21"
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Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>. | Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>. | ||
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== See also == | == See also == | ||
Revision as of 01:50, 9 December 2018
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. 
