Difference between revisions of "2015 AIME II Problems/Problem 10"
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Thus the number of permutations with n elements is three times the number of permutations with <math>n-1</math> elements. | Thus the number of permutations with n elements is three times the number of permutations with <math>n-1</math> elements. | ||
| − | However, for <math>n= | + | However, for <math>n=2</math>, there's an exception: there's only 2 places the 2 can go (before or after the 1). |
For <math>n=2</math>, there are <math>2</math> permutations. Thus for <math>n=7</math> there are <math>2*3^5=\boxed{486}</math> permutations. | For <math>n=2</math>, there are <math>2</math> permutations. Thus for <math>n=7</math> there are <math>2*3^5=\boxed{486}</math> permutations. | ||
Revision as of 20:39, 9 March 2017
Problem
Call a permutation 
of the integers 
quasi-increasing if 
for each 
. For example, 53421 and 14253 are quasi-increasing permutations of the integers 
, but 45123 is not. Find the number of quasi-increasing permutations of the integers 
.
Solution
The simple recurrence can be found.
When inserting an integer 
into a string with 
integers, we notice that the integer has 3 spots where it can go: before , before 
, and at the very end.
EXAMPLE: Putting 4 into the string 123: 4 can go before the 2: 1423, Before the 3: 1243, And at the very end: 1234.
Thus the number of permutations with n elements is three times the number of permutations with 
elements.
However, for 
, there's an exception: there's only 2 places the 2 can go (before or after the 1).
For , there are 
permutations. Thus for 
there are 
permutations.
See also
| 2015 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
