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Difference between revisions of "2015 AIME II Problems/Problem 10"

m (→‎Solution: Since recurrence is rather rare, I have decided to include a few words and a few other minor edits.)
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When inserting an integer <math>n</math> into a string with <math>n - 1</math> integers, we notice that the integer <math>n</math> has 3 spots where it can go: before <math>n - 1</math>, before <math>n - 2</math>, and at the very end.
 
When inserting an integer <math>n</math> into a string with <math>n - 1</math> integers, we notice that the integer <math>n</math> has 3 spots where it can go: before <math>n - 1</math>, before <math>n - 2</math>, and at the very end.
  
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EXAMPLE:
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Ex. Inserting 4 into the string 123:
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Putting 4 into the string 123:
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4 can go before the 2 (1423), before the 3 (1243), and at the very end (1234).
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4 can go before the 2: 1423,
 
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Before the 3: 1243,
 
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And at the very end: 1234.
 
  
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Only the addition of the next number, n, will change anything.
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Only the addition of the next number, <math>n</math>, will change anything.
  
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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.  
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Thus the number of permutations with <math>n</math> elements is three times the number of permutations with <math>n-1</math> elements.  
  
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However, for <math>n=2</math>, there's an exception: there's only 2 places the 2 can go (before or after the 1).
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Start with <math>n=3</math> since all <math>6</math> permutations work. And go up: <math>18, 54, 162, 486</math>.
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For <math>n=2</math>, there are <math>2</math> permutations. Now all we need to do is simple multiplication! For 1 through 7: 1, 2, 6, 18, 54, 162, 162*3=486.
 
  
 
Thus for <math>n=7</math> there are <math>2*3^5=\boxed{486}</math> permutations.
 
Thus for <math>n=7</math> there are <math>2*3^5=\boxed{486}</math> permutations.
  
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Note that recurrence is rather common when you see something of an ascending sequence BUT it has a condition such as "the next number can be 2 smaller"; this same idea appeared on another AIME with 8 boxes.
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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==
 
==See also==
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{{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}}
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[[Category: Intermediate Combinatorics Problems]]

Latest revision as of 11:15, 25 June 2024

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.

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, $n$, 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 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

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