2015 AIME II Problems/Problem 10

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Problem

Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$. For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$, but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$.

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.

Only the addition of the next number, n, will change anything.

Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.

Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$.

Thus for $n=7$ there are $2*3^5=\boxed{486}$ 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 a 8-box problem.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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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