Problem Solving

There are four ladies with different heights are posing for a photograph. Photographer want them to stand in 2 rows with shorter women in front row. How many arrangments are possible ?

The intent of the problem above is not crystal clear.
I suggest that instead you try the following problems, which test the same concept but are more clearly worded:

https://www.beatthegmat.com/permutations ... 28282.html
https://www.beatthegmat.com/combination-t205297.html

Caveat: The latter problem is a bit too involved for the GMAT.

If we take the problem to mean that no taller women should stand in front of a shorter woman, and that both rows are 'straight', and that when a person stands behind another they are aligned directly 'in tandem', and make no assumption that there must be an equal number in each row (as that is unrestricted in the question), and if a woman at the rear must have another standing directly in front, then there are 2 possible layouts:

A) 2 in the rear and 2 at the front
B) 1 in the rear and 3 at the front

No other layouts are possible because there must be 2 rows, and the front row contains women plural (not a woman).

I will label the women: 1,2,3,4 by rank order of height, smallest to largest.
Also, arrangements will be written in the form (back_row, front_row).

Layout type A (2x2)arrangements:

(43,21), (43,12), (34,21), (34,12), (42, 31), (24, 13) = 6

Layout type B:

(4,XYZ) has 6 arrangements of XYZ * 3 positions for the rear 4th lady = 24
(3,XYZ) has 6 arrangements of XYZ * 2 positions for the rear 3rd lady = 12
(2,XYZ) has 6 arrangements of XYZ * 1 position for the rear 2nd lady = 6

Total number of arrangements = 6 + 24 + 12 + 6 = 48

If we were to allow staggered arrangements whereby people could ONLY stand behind in the space between 2 others, then:

The number of Type A arrangements would be doubled to 12

and if a rear taller person must stand between two shorter front people, then
the number of type B arrangements would be found as follows:

(4,XYZ) has 6 arrangements of XYZ * 2 positions for the rear 4th lady = 12
(3,XYZ) has 2 arrangements of XYZ * 2 positions for the rear 3rd lady = 4

So the total number of staggered arrangements = 12 + 12 + 4 = 28

If we allow all of the above arrangements (aligned or staggered), then
N = 48 + 28 = 76

Here are the shapes of these arrangements:

R = rear individual
F = front individual
(ignore any dots, which are space holders)

Type A:

RR
FF

R R
.F F

.R R
F F

Type B:

R
FFF

. R
FFF

.. R
FFF

.R
F F F

.. R
F F F

Arrangements excluded:

R
.F F F

.. R
F F F

RR
.RR

R.. R
.RR

.RR
R.. R

All the above ensure that no woman is separated from the group, or else we might have no end to the answer.

After this labour-intensive journey, it is evident that the question needs to be more specific.
Nonetheless, it is interesting to see that with as few as only 4 women, there are so many possible arrangements. If we include the excluded arrangements, I expect we could exceed 100. With sitting and standing combinations, it would be considerably more.

That said, however, if the question answer options were limited as such:
(A) between 4 and 8 inclusive
(B) between 12 and 18 inclusive
(C) between 24 and 36 inclusive
(D) between 48 and 76 inclusive
(E) between 480 and 760 inclusive
then I would accept answer D, but I agree that it would not be like a GMat question.