combination problem

Solution:
A,B, C have to be before F, G, H.
Now D and E can occupy any space between each of A, B, C, F, G and H .
D and E can be case (1) together
Or case (2) separated by one other person.

Case (1) There are 7 places between A, B, C, F, G and H where they can be put in 7P1 * 2 ways. (multiplying by 2 because for every DE arrangement, there is an ED arrangement)
Besides A, B and C can be arranged among themselves in 3! Ways.
D, E and F can also be arranged among themselves in 3! ways.
So total ways of arranging A, B, C, D, E, F, G, and H such that A, B and C are always before F, G and H and D, E are together is 7P1*2*3!*3! = 504

Case (2) There are 7 places between A, B, C, F, G, and H where they can be put in 7P2 ways.
Besides A, B and C can be arranged among themselves in 3! Ways.
D, E and F can also be arranged among themselves in 3! ways.
So total ways of arranging A, B, C, D, E, F, G, and H such that A, B and C are always before F, G and H and D, E are not together is 7P2*3!*3! = 1512

So combining both cases correct answer is 1512 + 504 = 2016.

Taniuca wrote: I have tried first to get the totals by grouping ABC and FGH as one unit so that I can have 4 elements to be 4!=24
Then, I have found the number of times DE can be which is 2!=2
ABC= 3!=6 and FGH=3!=6 Why the answer cannot be 24*6*6*2 =1728

You are actually answering a slightly different question from the one that's asked. When you 'group ABC as one unit', you are then assuming that A, B and C are together in line, in some order. That is, you wouldn't be counting certain valid lineups, like ABDCEFGH, in which A, B and C do not appear in a row. So you haven't counted those lineups when either D or E is somewhere in the middle of the group ABC or the group FGH.

We can instead answer the question as follows: there are 8 places to put D, and then 7 places to put E. Once we've placed D and E, we know that A, B and C must go in the first three empty places, in some order; there are 3! orders we can put them in. Then in the last three empty spots, we must put F, G and H, and there are 3! orders to put them. So the answer is 8*7*3!*3! = 2016.