I am an architect working on a master plan for a campus of buildings. The total population of the campus is limited by parking capacity to 1,200 occupants. The campus has 6 floor plates to consider. We have developed 3 renovation concepts for the campus: Low-density, medium-density, and high-density. I want to determine how many ways the density options can be intermixed floor-by-floor, without exceeding the 1,200 occupant limit. The floors are not identical, so there is a total of 18 possible values, arranged in 6 sets of 3.
Floor 1: (140, 165, 179)
Floor 2: (120, 173, 215)
Floor 3: (119, 177, 218)
Floor 4: (128, 181, 214)
Floor 5: (87, 92, 150)
Floor 6: (187, 187, 278)
How many permutations of 6 values (1 from each set) yield a sum less than or equal to 1,200?
A computer could knock this out pretty easily, but a little bit of thought can solve it pretty easily with enumeration. First note that there are $729=3^6$ possible permutations. Taking the largest option in all cases yields a total of 1,254 occupants, which doesn't work but is only slightly over the limit.
The difference between the high and middle options in the six cases are: 14, 42, 41, 33, 58, and 91. The latter two differences are greater than 54, so any other permutation with more than 1,200 occupants must use the high density option for both the last two cases. But this shows that if we use the high option for all but one case, and the middle option for the other, we are overcapacity in the following cases: MHHHHH, HMHHHHH, HHMHHH, HHHMHH. In the same situation where we use 5 highs and a low, a similar analysis shows only the case LHHHHH is overcapacity.
If we allow two non-high options to be chosen, then the only way we can remain over capacity is if floors 1 and 4 are the chosen pair: choosing MHHMHH gives a total of 1,207 occupants. Lowering any of these options loses more than 7 occupants, so there are no other overcapacity permutations.
Of the 729 possibilities, only 7 are over capacity, so there are 722 permissible options.