solving by stars and bars methods 
In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?

Is there any way to solve this combination by stars and bars methods?  
I have got another method to solve this:

first d,e can come at any point,=>8c2 * 2 next all three a,b,c should
  come before f,g,h so take all a,b,c as one unit and f,g,h as one unit.
  each of which can be written in 3! ways, hence total no of ways of
  arrangement are 8c2*2*3!*3!=2016

But i have not got their explanation. Can you please explain this details.
 A: Firstly, "stars and bars" methods are usually used in situations where there are multiple copies of the same object being ordered. You have eight distinct objects (people) so it is unlikely that stars and bars will help.
The best strategy is to create a story of how the arrangement is created in time order. Each stage of the process has a certain number of choices, and you can multiply these together. You need to try to make sure that the number of choices at each stage is not affected by the choice you just made, otherwise you have to split into cases.
(Note that with "stars and bars", the stars and bars are a instructions in a process for creating a combination: the stars represent choosing a new object from the current type, and the bars represent switching to a new type of object.)
In this case, you could imagine the line being constructed one position at a time. F, G and H can't be in the first position, so you have 5 choices for who goes first. Then F, G and H still can't be in the second position, and you've chosen someone already, so there are 4 choices for second position. Similarly there are three choices for the next position. But now we run into problems because the number of choices now is dependent on whether all three of A, B and C have been chosen and we haven't specified that yet. So this is probably not the best story.
So how about we imagine constructing the line by letting each person choose where to stand in the line of 8? Since there are no restrictions on where D and E stand, we can start with them.
D has 8 choices for where to stand.
After that E has 7 choices for where to stand.
Now A can't choose the last three positions, so A has 3 choices.
Once A has chosen, B has 2 choices.
Once A and B have chosen, C has 1 choice.
Finally there are 3 positions for F to choose, then 2 positions for G and 1 position left for H.
So the total number is $8 \times 7 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1 = 2016$.
The story you presented goes like this:
You will construct the line in three groups: first D and E, then A, B and C, and last F, G and H. each with two parts.
First, D and E will choose. They can take up any two of the 8 positions, and these two positions can be chosen in $^8C_2$ ways.
Once they choose which two positions, they need to choose which of them will go first, which can be done in 2 ways.
Now there are 6 positions left to stand, and A, B, C must take the front three. They need to choose what order to stand in, which can be done in $3!$ ways.
There are now 3 positions left to stand, and D, E, F must take them. They need to choose what order to stand in, which can be done in $3!$ ways. 
Multiplying these together gives a total of $^8C_2 \times 2 \times 3! \times 3! = 2016$.
