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2006 AMC 12A Problems/Problem 25

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Problem

How many non- empty subsets S of \{1,2,3,\ldots ,15\} have the following two properties?

(1) No two consecutive integers belong to S.

(2) If S contains k elements, then S contains no number less than k.

\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ }  405

Solution

This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:

How many ways are there to choose k elements from an ordered n element set without choosing two consecutive members?

Note that if n < 2k - 1, the answer will be 0. Otherwise, the k elements we choose define k + 1 boxes into which we can drop the n - k remaining elements, with the caveat that each of the middle k - 1 boxes must have at least one element. This is equivalent to dropping n - 2k + 1 elements into k + 1 boxes, where each box is allowed to be empty. And this is equivalent to arranging n - k + 1 objects, k of which are dividers, which we can do in \displaystyle F(n, k) = {n - k + 1 \choose k} ways.

Now, looking at our original question, we see that the thing we want to calculate is just F(15, 1) + F(14, 2) + F(13, 3) + F(12, 4) + F(11, 5) = {15\choose 1} + {13\choose2} + {11\choose 3} + {9\choose 4} + {7 \choo...

See also

2006 AMC 12A (ProblemsResources)
Preceded by
Problem 24
Followed by
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