We know that 5 is the greatest number, so we'll take two cases:
(Also note that since 1 is the smallest number, it will have to go in the middle box, no number is less than 1 in the list)
Case 1. 5 is on the leftmost side of this inequality.
Then we will get:
5 > __ > 1 < __ < __
Subcase 1.1: 4 is on the rightmost side:
5 > __ > 1 < __ < 4
Now 2 or 3 can be placed in each of those 2 boxes, so resulting in 2 for this subcase.
Subcase 1.2: 4 is next to 5:
5 > 4 > 1 < __ <__
Note that there is only 1 option for this subcase as we can only place 2 and 3 in only one way.
Hence there are 3 options for this case.
Case 2. 5 is on the rightmost side of this inequality:
This inequality "mirrors" itself so this will also result in 3 options as well.
Hence, there will be \(\boxed{6}\) ways to order the integers 1 through 5 in the boxes.