Solution:

Label the seven columns of forward (rightward) segments; a path with no back segment simply chooses one forward segment in each column. The numbers of choices are $2,2,4,4,4,2,2,$ giving $2^{10}$ paths.

Let $s_1,s_2,s_3$ be the three left-pointing back segments (in columns $2,4,6$). Analyzing which columns become forced once a back segment is traversed gives $2^8$ paths for each of $\{s_1\}$ and $\{s_3\},$ $2^6$ for $\{s_1,s_3\},$ $2^9$ for $\{s_2\},$ $2^7$ for each of $\{s_1,s_2\}$ and $\{s_2,s_3\},$ and $2^5$ for $\{s_1,s_2,s_3\}.$

Adding, $$2^{10}+2\cdot2^8+2^6+2^9+2\cdot2^7+2^5=2400.$$

Thus, the correct answer is E.