Combinatorics was never easy

For each village on one side, let us consider the set of villages on the other side it directly connects to.

Label the villages on one side from $v_1$ to $v_{15}$ and the other side from $u_1$ to $u_{20}$ .

Clearly, the highest numbered village $v_i$ connects to equal to the lowest numbered village $v_{i+1}$ connects to. If it is more, the two bridges will intersect. Hence it is less than or equal. If it is less, then if $v_i$ connects to $u_j$ and $v_{i+1}$ connects to $u_k$ with $j<k$ , $u_k$ is not indirectly connected to $v_i$ .

If a village $v_i$ directly connects to $u_a$ and $u_b$ , it is directly connected to $u_c$ where $a\leq c\leq b$ . This can be obtained from the fact that if $u_c$ is connected to something else, then it causes an intersection of bridges, hence, $u_c$ would connect to $v_i$ .

Hence, this "partitions" the villages on one side into the villages of the other side. Instead of "partitioning" the villages, we look at the bridges. Each bridge is associated to exactly one village $v_i$ . We partition all $20+15-1$ bridges into $15$ villages. By stars and bars or otherwise, we get the answer of ${(20-1)+(15-1)\choose(15-1)}$ .