This is a question best approached by taking one condition at a time, and equations will make things even worse, I think. (You're never obliged to use algebra: only use it if it simplifies the problem!)
Let's call our six people ABCDEF with heights such that A > B > C > D > E > F.
A is the tallest, so we'll start with him. There's nobody taller than A, so nobody can stand behind A: this means A must go in the back row. Then, since there's nobody taller than A, he has to go on the right. This gives us
_ _ A
_ _ _
Now let's consider F. There's nobody shorter than F, so using logic just like we used above, F must go in the front row on the far left.
_ _ A
F _ _
Now we have a few options to play with. Let's start by placing B. Since B is the tallest person left, we have TWO options for him: he can go in front of A, or to the left of A.
_ B A
F _ _
or
_ _ A
F _ B
Now we'll place the shortest remaining person, E. Much like B above, E only has two options: he can go behind F, or to the right of F. Combining this with the above deductions, we now have four arrangements.
E B A
F _ _
_ B A
F E _
E _ A
F _ B
_ _ A
F E B
Now let's look at our four options carefully. If C and D are in the SAME COLUMN, we can't choose their arrangement: C is taller, so he goes on the right. Likewise, if C and D are in the SAME ROW, we can't choose their arrangement: C is taller, so he goes in the back.
In the first, third, and fourth arrangements above, C and D are either in the same column or are in the same row, so those three arrangements are DONE, and we have THREE arrangements so far.
The only arrangement we can play with is the second one, which has two permutations:
C B A
F E D
or
D B A
F E C
So we have a total of FIVE arrangements: the three that were DONE in the previous step, and the two more we just discovered.
Not as bad as it looks, if you take it condition by condition!
)
