Rectangle ABCD is constructed in the coordinate plane parall

IMPORTANT:

First notice that, to construct this rectangle, the vertices will share several points.
For example, if the 4 vertices are at (2, 5), (2, -3), (9, 5) and (9, -3), then we get a rectangle.

Notice that there are only 2 different x-coordinates (2 and 9) and only 2 different y-coordinates (-3 and 5)

So, to create the desired rectangle, we need only choose 2 different x-coordinates and 2 different y-coordinates

So, let's take the task of creating rectangles and break it into STAGES

STAGE 1: Select the 2 x-coordinates
We can choose 2 values from the set {3, 4, 5, 6, 7, 8, 9, 10, and 11}
In other words, we must choose 2 of the 9 values in the set
Since the order in which we choose the numbers does not matter, we can use COMBINATIONS
We can select 2 number from 9 numbers in 9C2 ways (= 36 ways)

STAGE 2: Select the 2 y-coordinates
We can choose 2 values from the set {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
In other words, we must choose 2 of the 11 values in the set
We can select 2 number from 11 numbers in 11C2 ways (= 55 ways)

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a rectangle) in (36)(55) ways (= 1980 ways)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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We need to choose 2 numbers from the x domain [3,11], since they will form two lines parallel to the y axis. Similarly, we need two values from the y domain [-5,5] to form two values parallel to the x axis. There are 9 integers for x and there are 11 numbers for y.

Choose 2 from 9 for the sides parallel to y axis: 9C2
Choose 2 from 11 for the sides parallel to x axis: 11C2

Multiply to get the overall number which should give you C.

Sorry for the double post- but another easier way to think about this is as so: we can calculate the number of rectangles just as we have done for the original question you provided here.

Take any rectangle and you have two diagonals. Each diagonal divides the rectangle into two different right triangles. So taking the two diagonals into account, we can create 4 right triangles with each rectangle. So just multiply the number of rectangles by 4 to get the answer..

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Hope it helps.

consider rectangle to be ABCD

choosing A's X and Y co-ordinate >> 10c1*9c1
choosing B's X and Y co-ordinate >> 9c1.1 (because one coordinate is fixed)
choosing C's X and Y co-ordinate >> 1.8c1 (because one coordinate is fixed)
choosing D's X and Y co-ordinate >> 1.1 (Because both coordinate are fixed)

this comes equal to 7920. I think I am considering some of the cases twice or even four times. Please tell me what am I doing wrong ..

Hi stunn3r

You have two errors:

(1) How did you come up with 7920, because 10C1*9C1*9C1*8C1 # 7920.
Because there are 9 ways to choose X, and 11 ways to choose Y

(2) I assume your equations are correct.
But the question here is "how many RECTANGLE?" not "how many combination of A,B,C and D ==> 4 points create only 1 rectangle ==> You should divide 7290/4 = 1980

Hope it helps.

The way I solved this problem was by thinking about the points and lines on the graph.

I chose two points on the horizontal axis between 3 & 11 . Because it doesn't specify that the points are integers I considered EVERY point, including the first, making the choices (9)(9) you then multiply by 11 because there are 11 points on the y axis. (9)(9)(11)= 891

You do the same thing for the y axis. 11 possible points and 2 must be chosen considering also that there are 9 horizontal points. (11)(11)(9) = 1089

Added together. (9)(9)(11) + (11)(11)(9) = 1980

An interesting thing is that the answer is a little smaller than 1980 because the axis points cannot be the same on the graph... but since there are an infinite number of points between 3 & 11, the difference is negligible.

I am trying to solve it in a different fashion. Can someone tell why this is wrong?

Assume that there are four points, ABCD.

Selecting point A = 9C1 * 11C1
Selecting point B = 8C1 (because one point on the X axis has been taken up by point A, leaving 8 points)
Selecting point C = 10C1 (because one point on the Y axis has been taken up by point A, leaving 10 points)
Selecting point D = 1C1 (for satisfying rectangle condition, only one point will be possible)

This gives 7920 options.

How will this question be different if we have to make a quadrilateral or a square?

I know my answer is wrong since for making a quadrilateral, we need two points on both the axes = 11C2 * 9C2

@buenuel please shed some light if you have spare bandwidth as I am horribly stuck.­

­Used the following approach, could someone please explain why this is incorrect? 
- Vertex A = 9c1 x 11c1 = 99 options 
- Vertex D = Since x co-ordinate is fixed, only 10 options for the y co-ordinate 
- Vertex B = Since y co-ordinate is fixed, only 8 options for the x co-ordinate
- Vertex C = Since everything else is fixed, only 1 option 

Therefore 99x10x8=7920 ways a rectangle can be formed 

Thanks!