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This is not an easy question. Notice that we need rectangles, not simply quadrilaterals.
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18 sided polygon will have 18 vertices, using combination i.e. C( 18,4) = 3060 E
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Bunuel
How many rectangles can be formed by taking the four vertices of 18-sided regular polygon?

A. 16
B. 24
C. 32
D. 36
E. 3060

straight lines: C(18,2)
triangles: C(18,3)
quadrilaterals: C(18,4)=18!/4!14!=3060
rectangle (4 right angles): 9!/2!=36

imagine the polygon in half and choose any two-points, from that half;
now, the number of ways we can choose the other two-points from opposite half is 1;
so, 18-sided polygon's half has 9 vertices to choose any two-points from: 9C2
and the number of ways to select the opposite vertices: 1
total: 9C2*1=36

Ans (D)

Hi, can't this also be done by something like this:
(18*17)/4

18 because can select any of the 18 vertices and 17 because after selecting the vertex, one can choose any of the remaining 17 vertices and then multiply by 1 because for the remaining 2 vertices there is just one possibility.

Divide by 4 for repetition, please let me know where I'm going wrong
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For making a rectangle we need 4 points ( at 90 deg angle when connected)
so we have to divide the polygon with 9 points on one side ( say 1,2,3..9) and 9 on the other
Now, from this 9 points
first select point 1, the combinations can be (1,2) (1,3) (1,4) ....(1,9) -----> 8 sets of rectangle
Note that we have the corresponding other points in the other 9 points (10, 11, ... 18)
Total we have 8 rectangle

similarly, for point 2, the combinations can be (2,3) (2,4) ...(2,9) ------> 7 sets of rectangle
same way we will have for all the points up to point 8, for which we have 1 rectangle

8+7+...1 =36
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