Count 'em All 11!

Let's fill in the holes so that it becomes a perfectly rectangular grid.

From this note , we know that the number of quadrilaterals in a $17\times 18$ grid is ${18\choose2}\times{19\choose2}=\frac{17\times18\times18\times19}{4}=26163$

Now we need to remove the quadrilaterals that are not part of the original grid. To do this, let's split it into 3 cases:

Case 1: Quadrilaterals with its top-left corner in region A and its bottom-right corner in region B

This is trivial, all we need to do is first choose the top-left corner in A, there are $5\times8=40$ ways to do so, then we choose the bottom-right corner in B, there are $6\times3=18$ ways to do so.

Hence the number of quadrilaterals with its top-left corner in region A and its bottom-right corner in region B is $40\times18=720$

Case 2: Quadrilaterals with its top-left corner in region A but its bottom-right corner not in region B

Suppose we choose the top-left corner of the quadrilateral $i$ units away from the left boundary of the whole grid and $j$ units away from the top boundary of the whole grid.

Then, the number of ways of choosing the bottom-right corner of the quadrilateral (excluding those lying in region B) is $(17-i)(18-j)-6\times3$ .

Thus, the total number of quadrilaterals with its top-left corner in region A but its bottom-right corner not in region B is $\begin{aligned} &\displaystyle \sum_{j=0}^{7}\sum_{i=0}^{4}[(17-i)(18-j)-18] \\&=\sum_{j=0}^{7}\sum_{i=0}^{4}[(j-18)i-17j+288] \\&=\sum_{j=0}^{7}\left[\frac{4\times5}{2}(j-18)-5(17j-288)\right] \\&=\sum_{j=0}^{7}(10j-180-85j+1440) \\&=\sum_{j=0}^{7}(1260-75j) \\&=7980 \end{aligned}$

Case 3: Quadrilaterals with its top-left corner not in region A but its bottom-right corner in region B

Similarly, suppose we choose the bottom-right corner of the quadrilateral $i$ units away from the right boundary of the whole grid and $j$ units away from the bottom boundary of the whole grid.

Then, the number of ways of choosing the top-left corner of the quadrilateral (excluding those lying in region A) is $(17-i)(18-j)-5\times8$ .

Thus, the total number of quadrilaterals with its top-left corner not in region A but its bottom-right corner in region B is $\begin{aligned} &\displaystyle \sum_{j=0}^{2}\sum_{i=0}^{5}[(17-i)(18-j)-40] \\&=\sum_{j=0}^{2}\sum_{i=0}^{5}[(j-18)i-17j+266] \\&=\sum_{j=0}^{2}\left[\frac{5\times6}{2}(j-18)-6(17j-266)\right] \\&=\sum_{j=0}^{2}(15j-270-102j+1596) \\&=\sum_{j=0}^{2}(1324-87j) \\&=3717 \end{aligned}$

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Therefore, the number of quadrilaterals in the original grid is $26163-720-7980-3717=13746$ .