In how many ways can A, B, C, D, E, G and H be arranged in a line if A has to be before D, and H has to be after D? The original question was about seven sculptors (Adam,Brian,Collins,Dorothy,Evelyn,George, and Henrietta *1) each having to display their artwork for a week, and how many possible display schedules there could be. Therefore, I interpreted the question as "Adam does not necessarily have to be directly before Dorothy, and Henrietta does not necessarily have to be directly after Dorothy" . (i.e. if Adam gets the first week and Dorothy the third week, Adam is still before Dorothy so this schedule is permissible).
*1 but referred to from here on out just by their initials since the full names couldn't fit in the title of the question
The way I approached it was to say D has to be in positions 2 to 6 - because if D is in position 1, A cannot be before it, and if it is in position 7, H cannot be after it.
Then with D in position
2: $1 \cdot 1 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways
3: $5 \cdot 4 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1=480$ ways
4: $5 \cdot 4 \cdot 3 \cdot 1 \cdot 3 \cdot 2 \cdot 1=360$ ways
5: $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 2 \cdot 1=240$ ways
6: $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 1 \cdot 1=120$ ways
which gives a total of $1320$ ways.
My reasoning was e.g. with D in the 3rd spot, the first position has $5$ possible options (anything except D or H). Then the second position has $4$ options (everything except D or H, or the letter that has already been chosen). The third position can only be taken by D. The fourth position can then be taken by any of the $4$ remaining letters, one of which will naturally be H.
However, my friend wrote out a different solution by manually constructing the different arrangements, each time keeping the distance between A and D constant and then distributing the H, moving those arrangements to the right, and multiplying by the permutation for the other letters. He got an answer of $840$. Please see the attached image for his workings.

The correct approach, and advice on possible faults in our various methods, would be greatly appreciated!
 A: Alternative approach:
There are $7!$ ways of placing the $7$ people, if the constraint re ordering $A,D,H$ is ignored.
By symmetry, you will have the same number of orderings for
each of the $3!$ ways of ordering $A,D,H$.
That is, you would expect there to be the same number of sequences for the $7$ people, when  $A,D,H$ occurs in each of the following $(3!)$ orders:
$ADH, ~~AHD, ~~DAH, ~~DHA, ~~HAD, ~~HDA.$
Therefore, the effect of the constraint is simply to divide the total number of sequences by $(3!)$.
Therefore, the answer is $\displaystyle \frac{7!}{3!}.$
A: A much more straightforward method is as follows -
First place A, D and H in a line in that order. Now we have to place B, C, E and G in the line. Start with anyone, say B.
Then B has a choice of $4$ places between A, D and H
_A_D_H_
Then the next person will have a choice of $5$ places and so on.
That gives you a total number of arrangements as $4 \cdot 5 \cdot 6 \cdot 7 = 840$
A: Your friend's solution is correct.  However, it is possible to avoid the case work both you and your friend employed.
Method 1:  Let's correct your solution.
You correctly solved the cases when Dorothy is scheduled second and when she is scheduled sixth.  In the other cases, you have not handled the restrictions that Adam appears before Dorothy and Henrietta appears after Dorothy correctly.
Dorothy is in the third position:  There are two ways to place Adam in one of the first two positions and four ways to place Henrietta in one of the last four positions.  There are $4!$ ways to arrange the remaining four people, so you should have obtained
$$2 \cdot 4 \cdot 4! = 192$$
such arrangements.
Dorothy is in the fourth position:  There are three ways to place Adam in one of the first three positions and three ways to place Henrietta in one of the last three positions.  There are $4!$ ways to arrange the remaining four people, so you should have obtained
$$3 \cdot 3 \cdot 4! = 216$$
such arrangements.
Dorothy is in the fifth positions:  There are four ways to place Adam in one of the first four positions and two ways to place Henrietta in one of the last two positions.  There are $4!$ ways to arrange the remaining four people, so you should have obtained
$$4 \cdot 2 \cdot 4! = 192$$
such arrangements.  Notice that this case is symmetric with the one having Dorothy in the third position.
Since the five cases are mutually exclusive and disjoint, the number of admissible permutations is
$$120 + 192 + 216 + 192 + 120 = 840$$
Method 2:  We first place Adam, Dorothy, and Henrietta, then arrange the remaining four people.
There are seven positions to fill.  Choose which three of those positions will be occupied by Adam, Dorothy, and Henrietta, which can be done in $\binom{7}{3}$ ways.  There is only one way to arrange Adam, Dorothy, and Harriet in those positions since Adam must be scheduled before Dorothy and Dorothy must be scheduled before Henrietta.  The remaining four people can be scheduled in the remaining four positions in $4!$ ways.  Hence, there are
$$\binom{7}{3}4! = 840$$
admissible arrangements.
Method 3:  We first schedule the other four people, then place Adam, Dorothy, and Henrietta in the three remaining positions.
We have seven positions to fill. There are seven ways to place Brian, which leaves six ways to place Collins, five ways to place Evelyn, and four ways to place George.  The remaining three positions must be filled by Adam, Dorothy, and Henrietta.  There is only one way to arrange Adam, Dorothy, and Harriet in those positions since Adam must be scheduled before Dorothy and Dorothy must be scheduled before Henrietta.  Hence, the number of admissible arrangements is
$$7 \cdot 6 \cdot 5 \cdot 4 = 840$$
which can also be expressed in the form
$$P(7, 4) = \frac{7!}{(7 - 4)!} = \frac{7!}{3!} = 840$$
using permutations.
Method 3 uses the falling factorial.  The method posted by Math Lover uses the rising factorial.  The case work that you and your friend did can also be avoided by using the symmetry argument that user2661923 made.
