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

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EXAMPLE:  
 
EXAMPLE:  
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putting 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,
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4 can go before the 2: 1423,
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before the 3: 1243,
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Before the 3: 1243,
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and at the very end: 1234.
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And at the very end: 1234.
  
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Thus the number of permutations with n elements is three times the number of permutations with n-1 elements.  
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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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For n=3, there are 6 permutations. Thus for n=7 there are 6*3^4 = 486 permutations.
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For <math>n=3</math>, there are <math>6</math> permutations. Thus for <math>n=7</math> there are <math>6*3^4=\boxed{486}</math> permutations.
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==See also==
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{{AIME box|year=2015|n=II|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 10:21, 27 March 2015

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.

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

For $n=3$, there are $6$ permutations. Thus for $n=7$ there are $6*3^4=\boxed{486}$ 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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