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Confused

by shibsriz@gmail.com » Sun Aug 04, 2013 2:29 am
Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle

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by Brent@GMATPrepNow » Sun Aug 04, 2013 5:41 am
shibsriz@gmail.com wrote:Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle
Notice that, if the rectangle is parallel to the x- and y-axes, then the coordinates of the 4 vertices will be such that:
- 2 vertices share one of the x-coordinates
- 2 vertices share the other x-coordinate
- 2 vertices share one of the y-coordinates
- 2 vertices share the other y-coordinate
For example, the points (8, -2), (11, -2), (8, 4) and (11, 4) create a rectangle AND they meet the above criteria.
So, to create a rectangle, all we need to do is select two x-coordinates and two y-coordinates.

Okay, now my solution . . .

Take the task of building rectangles and break it into stages.

Stage 1: Choose the two x-coordinates
The x-coordinates must be selected from {3,4,5,6,7,8,9,10,11}
Since the order of the selections does not matter, we can use combinations.
We can select 2 coordinates from 9 coordinates in 9C2 ways (36 ways).

Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Choose the two y-coordinates
The y-coordinates must be selected from {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Since the order of the selections does not matter, we can use combinations.
We can select 2 coordinates from 11 coordinates in 11C2 ways (55 ways).

By the Fundamental Counting Principle (FCP) we can complete the 2 stages (and build a rectangle) in (36)(55) ways ([spoiler]= 1980 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by vishugogo » Sun Aug 04, 2013 6:32 am
Dear Brent,

I was solving in another way but got stuck

I decided to solve for each co-ordinates

For A selection could be done in 9*11 ways

For B selection could be done in 8*11 ways

Now what??

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by Brent@GMATPrepNow » Sun Aug 04, 2013 6:54 am
vishugogo wrote:Dear Brent,

I was solving in another way but got stuck

I decided to solve for each co-ordinates

For A selection could be done in 9*11 ways

For B selection could be done in 8*11 ways

Now what??
I like your approach.

If you let A and B be points on the rectangle's diagonal, then a unique rectangle is defined by these two diagonal points. For example, the points (8, -2) and (11, 4) define the entire rectangle with points (8, -2), (11, -2), (8, 4) and (11, 4)

HOWEVER, this approach gets troublesome when we have to later go back and see how many duplicate rectangles we have created.


In your solution, you select point A (one of the vertices).
As you suggest, this can be done in (9)(11) ways (99 ways)
If point B is to be the point diagonal to point A, the x- and y-coordinates of point B must be different from the x- and y-coordinates of point A.
So, we can select point B in (8)(10) ways (80 ways)

So, the total number of ways to select points A and B = (99)(80) = 7920
HOWEVER, we have inadvertently counted some rectangles more than once.

For example, if we interchange labels for points A and B, we still get the same rectangle, so we have counted the same rectangle twice.
Also if we switch labels for points A and B with the labels for points C and D we still get the same rectangle as well, so we have actually counted the same rectangle 4 times.

So, we must divide 7920 by 4 to get 1980

Cheers,
Brent
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by karishma315 » Tue Aug 23, 2016 5:22 am
Hi Brent ,

I had a small question.

Like in this question you have used Fundamental counting Principle,which says that if a work has two tasks and one task can be done in x ways and another task can be done in y ways ,then the total no of ways to accomplish the work is x*y ways.

Now my question is ,"by total no of ways " do we mean "total UNIQUE ways" because in the below question you have not taken any added steps to remove duplicate rectangles.

Thanks for the help.

Regards
Karishma Duggal

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by karishma315 » Tue Aug 23, 2016 5:24 am
Brent@GMATPrepNow wrote:
shibsriz@gmail.com wrote:Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle
Notice that, if the rectangle is parallel to the x- and y-axes, then the coordinates of the 4 vertices will be such that:
- 2 vertices share one of the x-coordinates
- 2 vertices share the other x-coordinate
- 2 vertices share one of the y-coordinates
- 2 vertices share the other y-coordinate
For example, the points (8, -2), (11, -2), (8, 4) and (11, 4) create a rectangle AND they meet the above criteria.
So, to create a rectangle, all we need to do is select two x-coordinates and two y-coordinates.

Okay, now my solution . . .

Take the task of building rectangles and break it into stages.

Stage 1: Choose the two x-coordinates
The x-coordinates must be selected from {3,4,5,6,7,8,9,10,11}
Since the order of the selections does not matter, we can use combinations.
We can select 2 coordinates from 9 coordinates in 9C2 ways (36 ways).

Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Choose the two y-coordinates
The y-coordinates must be selected from {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Since the order of the selections does not matter, we can use combinations.
We can select 2 coordinates from 11 coordinates in 11C2 ways (55 ways).

By the Fundamental Counting Principle (FCP) we can complete the 2 stages (and build a rectangle) in (36)(55) ways ([spoiler]= 1980 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775





Hi Brent ,

I had a small question.

Like in this question you have used Fundamental counting Principle,which says that if a work has two tasks and one task can be done in x ways and another task can be done in y ways ,then the total no of ways to accomplish the work is x*y ways.

Now my question is ,"by total no of ways " do we mean "total UNIQUE ways" because in the below question you have not taken any added steps to remove duplicate rectangles.

Thanks for the help.

Regards
Karishma Duggal

GMAT/MBA Expert

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by Brent@GMATPrepNow » Tue Aug 23, 2016 6:08 am
karishma315 wrote: Hi Brent ,

I had a small question.

Like in this question you have used Fundamental counting Principle,which says that if a work has two tasks and one task can be done in x ways and another task can be done in y ways ,then the total no of ways to accomplish the work is x*y ways.

Now my question is ,"by total no of ways " do we mean "total UNIQUE ways" because in the below question you have not taken any added steps to remove duplicate rectangles.

Thanks for the help.

Regards
Karishma Duggal
Hi Karishma,

There is no duplication with my solution. We select 2 x-coordinates and 2 y-coordinates. That's all.
Once we've selected the 4 values, there's only 1 possible rectangle that can be created.

For example, if we select 8 and 11 for the x-coordinates and we select -2 and 4 for the y-coordinates, the ONLY rectangle that can get created has vertices at(8, -2), (11, -2), (8, 4) and (11, 4)

Cheers,
Brent
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