So, you are saying since there are 120 total ways to arrange a five bit string with ABCDE, there would be 20 of these in which A is before C and C is before E? Can I then think of it as choosing the other two members D and B in 5 x 4 ways? Or is that in error? Thanks so much for your prompt help!
since there got 5 string, let this to be 5 empty space
_ _ _ _ _
given situation some thing look like A->C->E (A before C before E)
from this 5 empty space, choose 3 of this 5, therefore there have C(5,3)=10
ways to put A C E, this left B and D, after finish put A C E, there left 2 free space for B and D. Hence C(5,3)*2*1=20