Rectangle ABCD is constructed in the coordinate plane parallel to the x and yaxes. If the x and ycoordinates of each of the points are integers which satisfy 3 â‰¤ x â‰¤ 11 and 5 â‰¤ y â‰¤ 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.)
1. 396
2. 1260
3. 1980
4. 7920
5. 15840
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Notice that, if the rectangle is parallel to the x and yaxes, then the coordinates of the 4 vertices will be such that:ankitbagla wrote:Rectangle ABCD is constructed in the coordinate plane parallel to the x and yaxes. If the x and ycoordinates of each of the points are integers which satisfy 3 â‰¤ x â‰¤ 11 and 5 â‰¤ y â‰¤ 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.)
1. 396
2. 1260
3. 1980
4. 7920
5. 15840
 2 vertices share one of the xcoordinates
 2 vertices share the other xcoordinate
 2 vertices share one of the ycoordinates
 2 vertices share the other ycoordinate
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 xcoordinates and two ycoordinates.
Okay, now my solution . . .
Take the task of building rectangles and break it into stages.
Stage 1: Choose the two xcoordinates
The xcoordinates 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/gmatcounting?id=789
Stage 2: Choose the two ycoordinates
The ycoordinates 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 = C[/spoiler])
Cheers,
Brent
Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmatcounting?id=775
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Hi Brent,Brent@GMATPrepNow wrote:...
Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: https://www.gmatprepnow.com/module/gmatcounting?id=789
...
Thanks for this EXTREMELY useful tip!
Regards,
Vivek
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I'm glad you like it!mevicks wrote:Hi Brent,Brent@GMATPrepNow wrote:...
Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: https://www.gmatprepnow.com/module/gmatcounting?id=789
...
Thanks for this EXTREMELY useful tip!
Regards,
Vivek
Cheers,
Brent
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Very nice tip  but one question, how would this work for other shapes, say, a triangle or parallelogram?
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For other shapes, we may be able to use pieces of the strategy.
For example, if one side of a triangle were parallel to the xaxis, then the two vertices on that side of the triangle would share the same ycoordinate.
Here's an example: https://www.beatthegmat.com/og13coordi ... 57190.html
Cheers,
Brent
For example, if one side of a triangle were parallel to the xaxis, then the two vertices on that side of the triangle would share the same ycoordinate.
Here's an example: https://www.beatthegmat.com/og13coordi ... 57190.html
Cheers,
Brent
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Thanks for that.Brent@GMATPrepNow wrote:For other shapes, we may be able to use pieces of the strategy.
For example, if one side of a triangle were parallel to the xaxis, then the two vertices on that side of the triangle would share the same ycoordinate.
Here's an example: https://www.beatthegmat.com/og13coordi ... 57190.html
Cheers,
Brent
I am not sure why we can't follow the strategy you posted for the rectangle? In other words, why not get the probably (in no particular order) of picking two points along the x axis and y axis like we did with the triangle? Is it because a rectangle flipped upside down is the same shape while a triangle flipped might look different (for example, if the coordinates of the right angle were (0,0 vs. 0,8)?
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Yeah, that's essentially it. I don't think Brent was dismissing your suggestion; he was (wisely) refraining from endorsing this as a universally applicable one or twostep method. You always have to consider rotations and reflections and determine which shapes are unique  these sorts of problems get well beyond GMAT difficulty very quickly.Zach.J.Dragone wrote:Thanks for that.Brent@GMATPrepNow wrote:For other shapes, we may be able to use pieces of the strategy.
For example, if one side of a triangle were parallel to the xaxis, then the two vertices on that side of the triangle would share the same ycoordinate.
Here's an example: https://www.beatthegmat.com/og13coordi ... 57190.html
Cheers,
Brent
I am not sure why we can't follow the strategy you posted for the rectangle? In other words, why not get the probably (in no particular order) of picking two points along the x axis and y axis like we did with the triangle? Is it because a rectangle flipped upside down is the same shape while a triangle flipped might look different (for example, if the coordinates of the right angle were (0,0 vs. 0,8)?
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Quick Question: While did we not multiply 1980 with 4! as we did in the 3 consonants and 2 vowels question(https://www.beatthegmat.com/howmanywordst279187.html)?
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Here's an illustrative example to explain why.Paras_0111 wrote:Quick Question: While did we not multiply 1980 with 4! as we did in the 3 consonants and 2 vowels question(https://www.beatthegmat.com/howmanywordst279187.html)?
The points A(8, 2), B(11, 2), C(8, 4) and D(11, 4) create a rectangle.
The points B(8, 2), D(11, 2), A(8, 4) and C(11, 4) create the same rectangle.
Cheers,
Brent
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Very true. This is already a very difficult counting question as is. Most other geometric shapes would probably produce results way outside the scope of the GMAT.Matt@VeritasPrep wrote:Yeah, that's essentially it. I don't think Brent was dismissing your suggestion; he was (wisely) refraining from endorsing this as a universally applicable one or twostep method. You always have to consider rotations and reflections and determine which shapes are unique  these sorts of problems get well beyond GMAT difficulty very quickly.Zach.J.Dragone wrote:Thanks for that.Brent@GMATPrepNow wrote:For other shapes, we may be able to use pieces of the strategy.
For example, if one side of a triangle were parallel to the xaxis, then the two vertices on that side of the triangle would share the same ycoordinate.
Here's an example: https://www.beatthegmat.com/og13coordi ... 57190.html
Cheers,
Brent
I am not sure why we can't follow the strategy you posted for the rectangle? In other words, why not get the probably (in no particular order) of picking two points along the x axis and y axis like we did with the triangle? Is it because a rectangle flipped upside down is the same shape while a triangle flipped might look different (for example, if the coordinates of the right angle were (0,0 vs. 0,8)?
Correct me if I'm wrong, but for a triangle... It would be 99C3, and then subtract all the combinations that produce 3 collinear points, which would be extremly difficult in a grid of this size. Especially starting from a total possibility of 99x98x97/6.
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Any time you're even using a word like concurrent or collinear, you know you're outside of the realm of the GMAT!800_or_bust wrote:Correct me if I'm wrong, but for a triangle... It would be 99C3, and then subtract all the combinations that produce 3 collinear points, which would be extremly difficult in a grid of this size. Especially starting from a total possibility of 99x98x97/6.
I solved it this way:
Rectangle ABCD
Stage 1: Choosing point A. The grid for this point is 9*11= 99
Stage 2: Choosing Point B. As Point B lies on the same y coordinate, we have 1 value for y and 8 possible values for x as one value of x is already taken for point A. Point B can be chosen in 8*1= 8
Stage 3: Choosing Point C. As Point C lies on the same x coordinate, we have 1 value for x and 10 possible values for y as one value of y is already taken for point A. Point C can be chosen in 1*10= 10
Stage 4: Choosing Point D. After choosing points A,B, and C the shape locks down and we have only 1 way to choose point D.
So, Total ways of shaping the rectangle is 99*8*10*1= 7920
What is wrong of my approach?
Thanks
Rectangle ABCD
Stage 1: Choosing point A. The grid for this point is 9*11= 99
Stage 2: Choosing Point B. As Point B lies on the same y coordinate, we have 1 value for y and 8 possible values for x as one value of x is already taken for point A. Point B can be chosen in 8*1= 8
Stage 3: Choosing Point C. As Point C lies on the same x coordinate, we have 1 value for x and 10 possible values for y as one value of y is already taken for point A. Point C can be chosen in 1*10= 10
Stage 4: Choosing Point D. After choosing points A,B, and C the shape locks down and we have only 1 way to choose point D.
So, Total ways of shaping the rectangle is 99*8*10*1= 7920
What is wrong of my approach?
Thanks
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This approach allows for identical rectangles to be counted more than once.Zoser wrote:I solved it this way:
Rectangle ABCD
Stage 1: Choosing point A. The grid for this point is 9*11= 99
Stage 2: Choosing Point B. As Point B lies on the same y coordinate, we have 1 value for y and 8 possible values for x as one value of x is already taken for point A. Point B can be chosen in 8*1= 8
Stage 3: Choosing Point C. As Point C lies on the same x coordinate, we have 1 value for x and 10 possible values for y as one value of y is already taken for point A. Point C can be chosen in 1*10= 10
Stage 4: Choosing Point D. After choosing points A,B, and C the shape locks down and we have only 1 way to choose point D.
So, Total ways of shaping the rectangle is 99*8*10*1= 7920
What is wrong of my approach?
Thanks
For example, in your approach, we might select (4,3) for point A, then select (10,3) for point B, then (4,5) for point C and then (10, 5) for point D This gets counted as 1 outcome.
However, if we select (10, 5) for point A, then select (4,5) for point B, then (10,3) for point C and then (4, 3) for point D, this gets counted as a different outcome (which it isn't).
Cheers,
Brent
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Thanks Brent for the prompt reply!This approach allows for identical rectangles to be counted more than once.
For example, in your approach, we might select (4,3) for point A, then select (10,3) for point B, then (4,5) for point C and then (10, 5) for point D This gets counted as 1 outcome.
However, if we select (10, 5) for point A, then select (4,5) for point B, then (10,3) for point C and then (4, 3) for point D, this gets counted as a different outcome (which it isn't).
Cheers,
Brent
Can you tell me why in this question you somehow followed the same approach I did that led to multiple counting?
https://www.beatthegmat.com/og13coordi ... 57190.html