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Hope it helps.
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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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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

Excellent explanation,

pls explain the similarity between between the triangle question and this, hope you would have solved it....I ll mention the question num for making it easy

OG 13 page num 185 question num 228

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.

I used the same method to slove just wanted to know the distinction between the question in my preceding post and the the question in this thread.

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

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..

I was thinking on the same lines
Now another had the question mentioned how many different squares instead of rectangle than what will be our answer???????????

= 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

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.

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

Can you explain this part in detail-That is why we get 11*10*10*9

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