Difference between revisions of "2007 AMC 12A Problems/Problem 25"
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In general, the number of subsets of a set with <math>n</math> element and with no <math>k</math> consecutive numbers is <math>\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}</math>. | In general, the number of subsets of a set with <math>n</math> element and with no <math>k</math> consecutive numbers is <math>\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}</math>. | ||
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== See also == | == See also == |
Revision as of 12:36, 21 July 2016
Problem
Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of including the empty set, are spacy?
Solution
Solution 1
Let denote the number of spacy subsets of . We have .
The spacy subsets of can be divided into two groups:
- those not containing . Clearly .
- those containing . We have , since removing from any set in produces a spacy set with all elements at most equal to and each such spacy set can be constructed from exactly one spacy set in .
Hence,
From this recursion, we find that
1 | 2 | 3 | 4 | 6 | 9 | 13 | 19 | 28 | 41 | 60 | 88 | 129 |
Solution 2
Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents .
For subsets of size there must be dividers between the balls, leaving dividers to be be placed in spots between the balls. The number of way this can be done is .
Therefore, the number of spacy subsets is .
Solution 3
A shifting argument is also possible, and is similiar in spirit to Solution 2. Clearly we can have at most elements. Given any arrangment, we subract from the element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size subsets of the set of the first positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for :
In general, the number of subsets of a set with element and with no consecutive numbers is .
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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