2007 AMC 12A Problems/Problem 25

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

And so the answer is $(E)$ $129$ .

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 $\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = \boxed{129}$ .

Solution 3

A shifting argument is also possible, and is similar 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 $n$ elements and with no $k$ consecutive numbers is $\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}$ .

Solution 4 (Casework)

Let us consider each size of subset individually. Since each integer in the subset must be at least $3$ away from any other integer in the subset, the largest spacey subset contains $4$ elements.

First, it is clear that there is $1$ spacey set with $0$ elements in it, the empty set. Next, there are $12$ spacey subsets with $1$ element in them, one for each integer $1$ through $12$ .

Now, let us consider the spacey subsets with $2$ elements in them. If the smaller integer is $1$ , the larger integer is any of the $9$ integers from $4$ to $12$ . If the smaller integer is $2$ , the larger integer is any of the $8$ integers from $5$ to $12$ . This continues, up to a smaller integer of $9$ and $1$ choice for the larger integer, $12$ . This means that there are $9 + 8 + \cdots + 1 = 45$ spacey subsets with $2$ elements.

For spacey subsets with $3$ elements, we first consider the middle integer. The smallest such integer is $4$ , and it allows for $1$ possible value for the smaller integer ( $1$ ) and possible $6$ values for the larger integer ( $7$ through $12$ ), for a total of $1 \cdot 6 = 6$ possible subsets. The next middle integer, $5$ , allows for $2$ smaller integers and $5$ larger integers, and this pattern continues up until the middle integer of $9$ , which has $6$ values for the smaller integer and $1$ value for the larger integer. This means that there are $1 \cdot 6 + 2 \cdot 5 + \cdots + 6 \cdot 1 = 56$ spacey subsets with $3$ elements.

Lastly, there are $3$ main categories for spacey subsets with $4$ elements, defined by the difference between their smallest and largest values. The difference ranges from $9$ to $11$ . If it is $9$ , there is only $1$ set of places to put the two middle values ( $n + 3$ and $n + 6$ , where $n$ is the smallest value). Since there are $3$ possible sets of smallest and largest values, there are $1 \cdot 3 = 3$ sets in this category. If the difference is $10$ , there are now $3$ sets of places to put the two middle values ( $n + 3$ and $n + 6$ or $7$ , and $n + 4$ and $n + 7$ ). There are $2$ possible sets of smallest and largest values, so there are $3 \cdot 2 = 6$ sets in this category. Finally, if the difference is $11$ , there are $6$ possible sets of places to put the two middle values ( $n + 3$ and $n + 6$ , $7$ , or $8$ , $n + 4$ and $n + 7$ or $8$ , and $n + 5$ and $n + 8$ ) and one possible set of smallest and largest values, meaning that there are $6 \cdot 1 = 6$ sets in this category. Adding them up, there are $3 + 6 + 6 = 15$ spacey subsets with $4$ elements.

Adding these all up, we have a total of $1 + 12 + 45 + 56 + 15 = \boxed{\mathrm{(E)}\ 129}$ spacey subsets. ~emerald_block

2007 AMC 12A (Problems • Answer Key • Resources)
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