AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
Personal tools

2006 AMC 12A Problems/Problem 25

From AoPSWiki

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
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us