Solution ( courtesy Manhattan GMAT staff)BL BM BR
FL FM FR
("B" stands for "back", "F" stands for "front", "L" stands for "left", etc.)
Let's also assign "names" to each of the six people - 1 is the shortest, 2 is the next shortest, ... and 6 is the tallest.
Notice first that the only place where 6 can stand is in the BR position. A person standing in any of the other positions has to be shorter than at least one other person, and 6 isn't shorter than anybody.
By similar reasoning, we can see that the only place where 1 can stand is in the FL position. A person standing in any of the other positions has to be taller than at least one other person, and 1 isn't taller than anybody.
So we know that any possible arrangement will be of this form:
BL BM 6
1 FM FR
All we need to do is count possible ways of putting 2, 3, 4, and 5 in positions BL, BM, FM, and FR. In order to count possibilities, let's focus on who goes into the BL position. 1 and 6 are already fixed in their own positions. There's no way 5 could be in the BL position, because there would be no way to assign someone to BM such that the heights in the back row increased consistently from left to right. So we know that the person in the BL position has to be either 2, 3, or 4. We investigate each possibility in turn:
If 2 goes in the BL position, there are just two possibilities:
2 4 6
1 3 5
and
2 5 6
1 3 4
If 3 goes in the BL position, there are also two possibilities:
3 4 6
1 2 5
and
3 5 6
1 2 4
If 4 goes in the BL position, there is just one possible arrangement:
4 5 6
1 2 3
Counting these possibilities, we see that there are only 5 possible arrangements.