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.

#HelpMe! #MathProblem

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

How did you do 1-5?

Jo ONG - 8 years, 3 months ago

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 ${10 \choose 3} - 8 - 6 - 4 - 2$

Lokesh Sharma - 8 years, 3 months ago

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

Calvin Lin Staff - 8 years, 2 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

Lokesh Sharma - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$