Pigeon Hole: The row of numbered houses problem, why couldn't I include the mentioned ranges in my pigeon holes options?

Question

The problem is as follow:

A row of houses are randomly assigned distinct numbers between 1 and 50 (inclusive). How many houses must there be to insure that there are 5 houses numbered consecutively?

Its solution:

Split the numbers into 10 pigeonholes: 1-5, 6-10, 11-15, 16-20… There must be at least =41 “pigeons”=houses

my problem is that: why haven't we include the ranges (2-6, 3-7, 4-8, 5-9, ... 12-16 ..) within the ones mentioned in the solution? they're consecutive 5 numbers as well, so why not?

Answer 1

So, taking "there are five houses numbered consecutively", which is slightly ambiguous, to mean "there are houses which use five consecutive numbers":

The solution given certainly demonstrates that there is no way to avoid five consecutive numbers given $41$ houses, based on the strong pigeonhole principle, and a small demonstration (e.g. remove the smallest of each group) shows that $40$ houses are not sufficient.

The use of the non-overlapping "pigeonholes" that use all the available numbers means there is less complexity about the proof presented, and since it achieves the result there is no need for further assessment.

Answer 2

The solution requires two parts:

1. With forty houses (namely all except $5,10,15,\ldots$), it is possible to not have five consecutives
2. With more than forty houses, we will always have five consecutive numbers

It does not matter how we prove the second point if only we do it in a way that wprks. If we want to use the pigeon-hole principle, we better find a couple of disjoint sets as "holes" into which we place our houses/pigeons. It suggests itself to use sets of five consecutive numbers as holes. We could use sets such as $\{1,2,3,4,5\}$ and/or $\{2,3,4,5,6\}$ and/or $\{3,4,5,6,7\}$, but to achieve disjointness, we better take $\{1,2,3,4,5\}$, $\{6,7,8,9,10\}$, $\{11,12,13,14,15\}$ and so on up to $\{46,47,48,49,50\}$. Fortunately, this results in $10$ "holes" so that of $41$ "pigeons", there will be at least five entering the same hole and we are done.

Other five-sets are simply not helpful.