No. of rectangles

Question

The coordinates (x,y) of each corners of rectangle ABCD are such that x and are integers and satisfy the equations 1 < x < 6 and -3 < y < -3. The edges of the rectangle are parallel to the X and Y axes. How many distinct rectangles could be formed which would satisfy these requirements?

ronniecoleman · Answer

x coordinates have values: 1 < x < 6 y coordinates have values: -3 < y < -3. 5c2 * 5c2 choose any two points on y axis choose any two points on x axis

ifairo · Answer

x coordinates have values: 1 < x < 6

I am assuming there is a mistake in the range of y, and correct range should be
y coordinates have values: -3 < y < 3.

x can take 2, 3, 4, and 5 => so 4 possible values.
y can take -2, -1, 0, 1, and 2 => 5 possible values.

4C2 * 5C2 = 60 possible rectangles.

What is the source of this question? What is the answer there?

rajataga · Answer

5c2 * 5c2

Just a minor correction - 4c2 * 5c2
which is 60.

I got this. But when i did it in a slightly different way, i got a different answer.
Can you just point out the mistake.

Total no. of points in this area = 20

Point A can now take up any of these points.
Now, point B, assuming AB is parallel to x-axis, can take up any of the other values of x (y will be the same), barring the x co-ordinate of point A.
Therefore, no. of points possible for Point B = 4-1 = 3

Point C (BC is parallel to y axis), C can take up any point on the y-axis (x will be same as Point B) , except those taken up by points A and B.

Hence, possibilities for Point C = 5-1 = 4.

Hence, total number of triangles = 20 x 3 x 4 = 240.

rajataga · Answer

Now that i write this, i realize why this difference is occurring.

each of the points is being taken as unique.

Hence, points ABCD could be interchanged by points DABC, and that is being counted as a separate rectangle.

However, shouldn't this hold true?

would rectangle ABCD be the same as DABC, because the four points are basically the same, though in the xy plane, D is present instead of A.

rajataga · Answer

Also, would you consider y=0 as one of the possibilities???

Even though it falls in the given range, y=0 would essentially mean that that edge lies on the X axis itself, and hence not parallel to the X-axis as given in the question.

Brent@GMATPrepNow · Answer

Total no. of points in this area = 20

Point A can now take up any of these points.

Rajata's observation provides an interesting/differnet way to tackle this question.
We can reduce our calculations by recognizing that a rectangle in this question is defined by 2 points (rather than 4). All we need are the coordinates of 2 vertices that are diagonal from each other.
For example, the vertices at (2,0) and (5,3) can define only one rectangle.

So, this counting problem is reduced to selecting 2 points in the given region.
The first point can be selected in 20 ways.
Once we have selected the first point, we need to select another point which is diagonal to this point. If this is the case, our second point cannot have the same x- or y-coordinate as the first point. This means that, when selecting the second point, we have only 12 points from which to choose.
So, the total number of ways to select 2 points from the region is 20x12=120 ways.
BUT, we need to divide this by 2 since we have counted each pair of coordinates twice (e.g., selecting points (2,0 and 5,3) is the same as selecting points (5,3) and (2,0).

Brent Hanneson - Creator of GMATPrepNow.com

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