Explanation about solution for selecting $5$ out of $8$ integers and a pair adds to $9$

Question

I am reading the following exercise in a section about the pigeohole principle.

Show that if five integers are selected from the first 8 positive integers, there must be a pair of these integers with sum equal to $9 $

The solution states:

Group the first 8 positive integers into four subsets of two integers, each so that the integers of each subset add up to $9$ i.e. {$1$,$8$}, {$2$,$7$},{$3$,$6$},{$4$,$5$}. If five integers are selected by the first 8, then by the pigeonhole principle at least two of them come from the same subset

I don't understand the explanation. The $4$ subsets mentioned are just one way to divide the $8$ positive integers. Another way would be {$1,7$}, {$2$,$8$}, {$3$,$5$}, {$4$,$6$} and in that case the pigeonhole principle does not help.
So is the fact that there is a specific way to group the $8$ positive integers into $4$ groups and apply the pigeonhole principle a valid solution since it is not generic enough?

The way I read the exercise is that a random choice of $8$ positive integers leads to a pair that sums to $9$. So how is the fact that we can construct a specific way to group pairs (the way I read the solution) a correct solution if there are other ways to group them that don't sum to $9$? Doesn't the random selection imply any of the possible sub groupings is equally likely? Hence we could have {$1,7$}, {$2$,$8$}, {$3$,$5$}, {$4$,$6$}

Answer 1

You are missing the point.

Suppose that I try to choose $5$ numbers from $\{1,2,\cdots,8\}$ such that no two of the $5$ sum to $9$.

How should I do it?

One try is selecting the numbers $\{1,2,3,4,5\}$. However, this fails, because $4 + 5 = 9.$

So, I try to vary it. I try $\{1,2,3,4,6\}$. This fails because $3 + 6 = 9.$

Then, I try $\{1,2,4,6,8\}$. This fails because $1 + 8 = 9.$

---

Then, I ask myself: is there some deeper pattern that is preventing me from succeeding.

I consider the following pairs of numbers:
$\{1,8\}$
$\{2,7\}$
$\{3,6\}$
$\{4,5\}$.

Then, I look at my $3$ failed attempts and I realized that the reason that they failed is that each attempt happened to select more than one number from one of the $4$ sets of numbers above.

So, then I ask myself: what do I have to do to select $5$ numbers such that no two of the numbers add up to $9$. It is clear that I must not select more than one number from any of the $4$ sets of numbers above.

Then, the light dawns. I am required to select five numbers. If I only permit myself to select one number from each of the $4$ sets above, then I will have selected at most $4$ numbers.

This implies that when I go to select the $5$-th number, I will have no choice but to take the second number from one of the $4$ sets above.