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

*

*With forty houses (namely all except $5,10,15,\ldots$), it is possible to not have five consecutives

*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.
A: 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.
