Let the persons in the order of Ascending height be 1 2 3 4 5 6
2nd ROW D E F
1st ROW A B C where A to F are the six places
A has to taken by 1 - no one shorter to stand behind him/her of to his left.
F has to be taken by 6 . He cannot be in the first row no on shorter than him to stand behind him. In 2nd Row in D&E he will be have someone shorter than hi, to his left.
B - cannot be taken by 5 as he is the tallest now remember 6 is fixed in F and no one to stand behind. The logic is wrong , ignore this part. So no. of ways 3
C - cannot be taken by 2 as the person to his right has to be shorter than him. Only 1 is shorter and his position is fixed. So no of ways 3. So 1st ROW can be arranged in 3*3 - 9 ways.
IMO E

IMO A

[spoiler]
(1)
D E F
A B C

(2)
B E F
A C D

(3)
B D F
A C E

(4)
C D F
A B E


(5)
C E F
A B D
[/spoiler]

Any other arrangement violates the two rules within the problem.

Yes because even the second position can be taken by the 2 and 3. So B can be filled in 2 ways.
Similar problem has been discussed before.

A B C D E F be the 6 people with increasing heights.

A will always stand in the left corner of row 1 and F will always stand in the right corner of row 2.

so, we basically have 4 people ( B C D E) to arrange in 4 seats. 2 in the front and 2 in the back.

2 people from 4 people who will sit in the first row can be selected in 4C2 ways = 6.
but one of combination will be a case when D and E are in the first row. This is not correct . Atleast one of D or E in the must be in second row. hence subtract 1.
hence, answer = 6 - 1 =5