# No to Arithmetic Progression

I have question which I am trying to solve but I am unable to do so.

Taking three or more elements from the set N, where N = {1,2,3,...,9,10}, how many sets can you create such that no three numbers in the created set are in an Arithmetic Progression.

I solved the question where N = {1,2,3,4,5}, but this one is out of bounds.

Note by Amit Tigga
5 years, 4 months ago

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How about calculating the total number of sets possible and subtracting from it number of sets in which AP is formed. That might give the required number of sets.
Answer is probably this $${10 \choose 3} - 8 - 6 - 4 - 2$$

- 5 years, 4 months ago

This doesn't include sets of 4 or more elements, like $$\{1, 3, 6, 10\}$$.

Staff - 5 years, 3 months ago

I wish, the question were just to find sets containing three elements then my answer might have been correct. Anyhow, I have another thought. What if we treat the three elements which forms an AP like an element to be used to form sets of four elements. Let's name all those sets of three elements which I am treating like an element as bomb element. Now find all the four element sets containing bomb element and subtract that from the total number of four element sets. Then treating the four element set forming an AP as next bomb elements one can find the number of permissible five set elements and the thing goes on until you explode. Lol

- 5 years, 3 months ago

How did you do 1-5?

- 5 years, 4 months ago