2015 AIME II Problems/Problem 10
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.
Ex. Inserting 4 into the string 123: 4 can go before the 2 (1423), before the 3 (1243), and at the very end (1234).
Only the addition of the next number, , will change anything.
Thus the number of permutations with elements is three times the number of permutations with
elements.
Start with since all
permutations work. And go up:
.
Thus for there are
permutations.
When you are faced with a brain-fazing equation and combinatorics is part of the problem, use recursion! This same idea appeared on another AIME with an 8-box problem.
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.