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 $S_{n + 1}$ can be divided into the subsets containing $n + 1$ and the ones not containing $n + 1$ . The latter is just $S_{n}$ , whereas the former is $S_{n - 2}$ (since removing $n + 1$ from any of these sets produces a spacy set with maximal element $n - 2$ ). Hence,

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

From this recursion, we find that

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

As a last resort, we can brute force the result by repeated casework. Luckily, 12 is not a very large number, so solving it this way is still possible.

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