Permutations and Combinations with conditions Please I'm having difficulties solving this question:
In how many ways A, B, C, D, E, F, G, H can be arranged such that A, B, and C always come before F, G and H?
 A: First, there are $3!$ ways to arrange $\{A,B,C\}$, and $3!$ ways to arrange $\{F,G,H\}$, and we know that the former are to the left of the latter.
The last two letters, $D$ and $E$, are entirely free to occupy the seven spaces surrounding the six letters that are already placed. These free letters could share any of $7$ spaces, or they could occupy any of $\binom72$ pairs of spaces. In any case, they could be in one of two orders. Thus, we obtain:
$$3!\cdot 3!\cdot 2\left(7+\binom72 \right)=2016$$ 
Does that work?

An easier way to get the third factor: Once the six letters with conditions are placed, there are $7$ places for $D$, and once it's placed, there are $8$ places for $E$. The product $7\cdot8$ equals $56$, and it's a bit easier to deal with than $2\left(7+\binom72 \right)$, which is also $56$.
A: If you choose a permutation of the $8$ letters at random, focus only on the $6$ positions containing a letter from $\{A,B,C,F,G,H\}$, and consider the subset of the $3$ of those positions that contain $A,B,C$, then any one of the $\binom63=20$ possible subsets is equally likely to occur. In order for $A,B,C$ to all come before each of $F,G,H$ in the the original permutation, requires precisely that one particular subset is so obtained: the first $3$ out of $6$ positions. So the number of such permutations of the $8$ letters is $\frac{8!}{20}=2016$.
