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. | ||
− | For <math>n= | + | However, for <math>n=1</math>, there's an exception: there's only 2 places the 2 can go (before or after the 1). |
+ | |||
+ | For <math>n=1</math>, there are <math>2</math> permutations. Thus for <math>n=7</math> there are <math>2*3^5=\boxed{486}</math> permutations. | ||
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=9|num-a=11}} | {{AIME box|year=2015|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:13, 22 May 2015
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 n into a string with n-1 integers, we notice that the integer n has 3 spots where it can go: before n-1, before n-2, 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 |
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