As the rectangle is parallel to coordinate axes, the coordinates of the points of the rectangle would be

(X1, Y1), (X2, Y1), (X2, Y2), (X1,Y2)

given that X1, X2 lie between 3 and 11..ie., 9 possible numbers

Possible combinations for X1,X2 would be 9C2 = 36

Similarly, Possible combinations for Y1, Y2 would be 11C2 = 55

Possible ways of constructing rectangle is by selecting any of the combination of X1,X2 and Y1,Y2

= 36 * 55 = 1980
Ans. C

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.

Just had a look- it is rather subtle. Here, we can choose 2 values of x and y as taking any 2 values of x and y will always yield a rectangle. For instance x=2 and x=3 are two lines parallel to y axis and y=1 and y=4 are two values parallel to x axis- plot these lines and you get a rectangle.

For the triangles query, we will need to constrain the values of the coordinates that x and y can take, say P is (x1,y1), Q(x1,y2) and R(x2,y1).

So we will need to pick one x to designate x1 and then we will have 9 more x values remaining from which we choose x2. Similarly we can do the same for y1 and y2.

That is why we get 11*10*10*9

If we choose 2 values directly , then we do not make a distinction in the order and if this happens we multiply the value by 4. Take x1=5, x2=8, y1=7 and y2=9 (so 8,5,7 and 9 can be arranged among one another and the combination does not take the changing values into account)

the four triangles you get are: (5,7), (8,7), (5,9) ; (8,7), (8,9), (5,7) ; (5,9), (8,9), (5,7) ; (8,9) , (5,9) , (5,7).

So we can also use : 4* 11C2*10C2 to get 9900

For the rectangle, choosing 2 values of x and y result in only 1 rectangle.
This is the only difference

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Hope it helps.
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= Select 2 points along the x-axis * Select 2 points along the y-axis
= \(\frac{9!}{2!7!} * \frac{11!}{2!9!} = 1980\)

Answer: 1980
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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.
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hi experts please tell me what is wrong with it

9*11(no of ways to select first point) *8 (another point on x axis) * 10(no of ways for another point on y axis)

BTW OA is 9*11(no of ways to select first point) *8 (another point on x axis) * 10(no of ways for another point on y axis) / 2

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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