2015 AIME II Problem 10

Below is the professionally curated solution for Problem 10 of the 2015 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AIME II solutions, or check the answer key.

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Concepts:permutationsrecursive counting

Difficulty rating: 2890

10.

Call a permutation a1,a2,,ana_1, a_2, \ldots, a_n of the integers 1,2,,n1, 2, \ldots, n quasi-increasing if akak+1+2a_k \le a_{k+1} + 2 for each 1kn1.1 \le k \le n - 1. For example, 5342153421 and 1425314253 are quasi-increasing permutations of the integers 1,2,3,4,5,1, 2, 3, 4, 5, but 4512345123 is not. Find the number of quasi-increasing permutations of the integers 1,2,,7.1, 2, \ldots, 7.

Solution:

Let SnS_n be the number of quasi-increasing permutations of 1,,n.1, \ldots, n. Insert nn into a quasi-increasing permutation of 1,,n1:1, \ldots, n - 1: the entry following nn must be at least n2,n - 2, so nn can go immediately before n1,n - 1, immediately before n2,n - 2, or at the very end — exactly 33 positions, and each insertion keeps every other adjacent condition intact.

Conversely, deleting nn from a quasi-increasing permutation of 1,,n1, \ldots, n leaves a quasi-increasing permutation of 1,,n1,1, \ldots, n - 1, since the entries around the deleted nn satisfy ak1n1ak+1+2a_{k-1} \le n - 1 \le a_{k+1} + 2 when n3.n \ge 3. So Sn=3Sn1S_n = 3S_{n-1} for n3.n \ge 3.

Since S2=2,S_2 = 2, we get S7=235=486.S_7 = 2 \cdot 3^5 = 486.

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