Difference between revisions of "2007 AMC 12A Problems/Problem 25"

(Solution)
(Solution 3)
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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>.
 
Solution 4
 
 
For each positive integer <math>n</math>, let <math>S_n = \{k:1\leq k\leq
 
n\}</math>, and let <math>c_n</math> be the number of spacy subsets of <math>S_n</math>. Then <math>c_1=2</math>, <math>c_2=3</math>, and <math>c_3=4</math>. For <math>n\geq 4</math>, the spacy subsets of <math>S_n</math> can be partitioned into two types: those that contain <math>n</math> and those that do not. Those that do not contain <math>n</math> are precisely the spacy subsets of <math>S_{n-1}</math>. Those that contain <math>n</math> do not contain either <math>n-1</math> or <math>n-2</math> and are therefore in one-to-one correspondence with the spacy subsets of <math>S_{n-3}</math>. It follows that <math>c_n=c_{n-3}+c_{n-1}</math>. Thus the first twelve terms in the sequence <math>\left(c_n\right)</math> are <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, <math>9</math>, <math>13</math>, <math>19</math>, <math>28</math>, <math>41</math>, <math>60</math>, <math>88</math>, <math>129</math>, and there are <math>c_{12}=\boxed{129}</math> spacy subsets of <math>S_{12}</math>.
 
  
 
== 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 $\{1,2,3,\ldots,12\},$ including the empty set, are spacy?

\[\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129\]

Solution

Solution 1

Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$.

The spacy subsets of $S_{n + 1}$ can be divided into two groups:

  • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$.
  • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$.

Hence,

$S_{n + 1} = S_{n} + S_{n - 2}$

From this recursion, we find that

$S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$
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 $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ 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 ${1,4,7,10}$.

For subsets of size $k$ there must be $2(k - 1)$ dividers between the balls, leaving $12 - k - 2(k - 1) = 12 - 3k + 2$ dividers to be be placed in $k + 1$ spots between the balls. The number of way this can be done is $\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k$.

Therefore, the number of spacy subsets is $\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = 129$.

Solution 3

A shifting argument is also possible, and is similiar in spirit to Solution 2. Clearly we can have at most $4$ elements. Given any arrangment, we subract $2i-2$ from the $i-th$ element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size $k$ subsets of the set of the first $14-2k$ 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 $k$: ${14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}$

In general, the number of subsets of a set with $n$ element and with no $k$ consecutive numbers is $\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}$.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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