Arithmetic progression in $[1,8]$

Question

Let we have the set $\{1,2,3,4,5,6,7,8\}$ and we take any $6$ numbers from this set, i.e. this can be done by $\tbinom{8}{6}=28$ ways. Prove that each of these $6$-subsets contains 3-term arithmetic progression.

(!) FALSE: I believe that this is true because any $6$-subsets of the set $\{1,2,3,4,5,6,7,8\}$ contains three consecutive numbers, i.e. $n,n+1,n+2$.

But I am not able to prove it rigorously.

Would be very grateful if anyone will help me.

EDIT: The user Arthur showed that my approach is false sine we can take $\{1,2,4,5,7,8\}$ which does not contain three consecutive integers. However, how to prove the initial statement?

Answer 1

Hint:

It is equivalent to removing two elements from the eight element set.

Now, these are the cases:

• The two elements are together. (in this case you'll always have three consecutive elements (an Arithmetic Progression) in your new set)

• The two elements are separated.

  • When the two elements are separated by a gap = $6$ or $5$ or $4$ or$3$, there are always three consecutive integers remaining between them .

    When they are separated by a gap = $2$, there are only two cases possible:

    1) gap of $2$ between $2$ elements and $2$ other elements, you obtain an AP with common difference = $3$.

    2) gap of $2$ between $1$ element and $3$ other elements. (again consecutive $3$ elements)

Prove that in each case (you just need to think about the second case now), an AP remains in your leftover group.

Answer 2

Consider $m$-th and $n$-th numbers are not chosen ($m<n$). When $n-m=1;2;4;5;6;7$, there are three consecutive numbers. When $n-m=3$, there are three alternative consecutive (difference two) numbers.