Topological Ordering
Unfortunately, there is no "fast" way to count them generally as this problem is #P-Complete. This means that it is basically as hard as every counting version of NP-problems, and is likely not solvable in polynomial time. 

For small examples such as this one you can come up with some ad-hoc counting technique however. In this case would try listing all permutations of E,F,G and see how many choices I have for the rest if I insert them in some order. Like this:


EFG: 1+2+4+6 = 13
(1 choice for A, then 2 for B, 4 for C, 6 for D)
EGF: 1+2+4+5 = 12 (in A,B,C,D order)
FEG: 1+2+4+5 = 12 (in B,C,A,D order)
FGE: 1+2+4+6 = 13 (in B,C,D,A order)
GEF: 1+2+4+5 = 12 (in D,C,A,B order)
GFE: 1+2+4+5 = 12 (in D,C,B,A order)


So in total 4*12 + 2*13 = 74 topological orderings (with some non-negligeable probability that I miscounted somewhere).