Combinatorics intense question

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The discussion revolves around finding the number of non-empty subsets of the set {1,2,3,...,15} that meet two criteria: no two consecutive integers can be included, and if a subset contains K elements, it cannot include any number less than K. Participants express confusion over the combinatorial approach, particularly the expression "n-k+1 choose k" and its implications. There is a mention of using triangle numbers to solve the problem, indicating a potential connection to combinatorial identities. Overall, the thread highlights the complexity of the problem and the need for a clearer understanding of combinatorial principles.
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Homework Statement


How many non- empty subsets of {1,2,3...,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 .



Homework Equations



choosing identities, not sure which one

The Attempt at a Solution


Tried casework, but I don't understand how the solution gets
n-k+1 choose k
 
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you said n-k+1 choose k
what is n?
 
If we have ways are there to choose k elements from an ordered n element set without choosing two consecutive members
 
newchie said:
If we have ways are there to choose k elements from an ordered n element set without choosing two consecutive members

Okay, so what does it mean when you take a choose b?
You're taking the number of ways to choose b items out of a, right?

So why would you suppose the answer to your question is n-k+1 choose k?
What is the significance of n-k+1?
 
http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems/Problem_25
I don't understand the solution
 
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What about it do you not understand?
 
Defining the boxes into n-2k+1 elements, then the general choosing statement, i get the first part, but the second part is a bit opaque
 
Interesting problem! Fiddling with pencil & paper and looking for patterns suggests that for n integers, the answer involves sums of triangle numbers, T(n) = n(n+1)/2, but I don't see a good way to get there from simple combinetrics.
 
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